How Do You Know if a Compound Will Form a Nuetral Solution in Water

Chapter 14. Acrid-Base of operations Equilibria

xiv.4 Hydrolysis of Salt Solutions

Learning Objectives

Past the terminate of this section, y'all will be able to:

• Predict whether a common salt solution will exist acidic, basic, or neutral
• Calculate the concentrations of the various species in a salt solution
• Describe the procedure that causes solutions of sure metallic ions to be acidic

Every bit we accept seen in the section on chemic reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word "neutralization" seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall at present see.

Acid-Base of operations Neutralization

A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When nosotros mix solutions of an acrid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, nosotros may discover that the resulting solution is non neutral. Information technology could comprise either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or bones. The following four situations illustrate how solutions with various pH values can ascend following a neutralization reaction using stoichiometrically equivalent quantities:

1. A stiff acid and a potent base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the cohabit partners produced are of negligible strength (see Chapter xiv.3 Relative Strengths of Acids and Bases):

   [latex]\text{HCl}(aq)\;+\;\text{NaOH}(aq)\;{\rightleftharpoons}\;\text{NaCl}(aq)\;+\;\text{H}_2\text{O}(l)[/latex]

2. A strong acid and a weak base yield a weakly acidic solution, non because of the potent acid involved, just because of the conjugate acid of the weak base.
3. A weak acid and a stiff base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of operations of the weak acid and the conjugate acid of the strong base. The cohabit acid of the strong base is a weaker acid than water and has no event on the acerbity of the resulting solution. Nonetheless, the conjugate base of the weak acid is a weak base of operations and ionizes slightly in h2o. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.
4. A weak acid plus a weak base can yield either an acidic, bones, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can exist either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acrid and the weak base volition have the same strength, and then their respective conjugate base and acid will take the aforementioned strength, and the solution will be neutral. To predict whether a detail combination will exist acidic, basic or neutral, tabulated 1000 values of the conjugates must exist compared.

Stomach Antacids

Our stomachs incorporate a solution of roughly 0.03 M HCl, which helps us digest the food nosotros consume. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the tum is, and the results tin can be very painful. When we have heartburn, information technology feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the about common antacids is calcium carbonate, CaCO₃. The reaction,

[latex]\text{CaCO}_3(southward)\;+\;2\text{HCl}(aq)\;{\rightleftharpoons}\;\text{CaCl}_2(aq)\;+\;\text{H}_2\text{O}(50)\;+\;\text{CO}_2(m)[/latex]

non only neutralizes stomach acid, it also produces CO_ii(g), which may result in a satisfying belch.

Milk of Magnesia is a break of the sparingly soluble base magnesium hydroxide, Mg(OH)_two. It works according to the reaction:

[latex]\text{Mg(OH)}_2(southward)\;{\rightleftharpoons}\;\text{Mg}^{ii+}(aq)\;+\;2\text{OH}^{-}(aq)[/latex]

The hydroxide ions generated in this equilibrium and so go on to react with the hydronium ions from the breadbasket acrid, so that :

[latex]\text{H}_3\text{O}^{+}\;+\;\text{OH}^{-}\;{\rightleftharpoons}\;2\text{H}_2\text{O}(l)[/latex]

This reaction does not produce carbon dioxide, only magnesium-containing antacids can have a laxative effect.

Several antacids have aluminum hydroxide, Al(OH)_iii, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to residue the side effects of the two substances.

Culinary Aspects of Chemical science

Cooking is essentially synthetic chemistry that happens to exist prophylactic to swallow. There are a number of examples of acid-base of operations chemistry in the culinary world. One case is the employ of blistering soda, or sodium bicarbonate in baking. NaHCO₃ is a base of operations. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbling of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter "rises." Baking pulverisation is a combination of sodium bicarbonate, and ane or more acid salts that react when the two chemicals come in contact with water in the concoction.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure ane). Information technology turns out that fish have volatile amines (bases) in their systems, which are neutralized past the acids to yield involatile ammonium salts. This reduces the scent of the fish, and as well adds a "sour" taste that we seem to enjoy.

Effigy 1. A neutralization reaction takes place between citric acid in lemons or acerb acid in vinegar, and the bases in the flesh of fish.

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste matter product in a process called fermentation. The lactic acrid somewhen increases the acidity of the brine to a level that kills whatsoever harmful bacteria, which crave a basic environment. Without the harmful leaner consuming the cucumbers they are able to concluding much longer than if they were unprotected. A byproduct of the pickling procedure changes the flavour of the vegetables with the acid making them gustatory modality sour.

Salts of Weak Bases and Stiff Acids

When we neutralize a weak base with a strong acrid, the product is a salt containing the cohabit acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH₄Cl, is a salt formed past the reaction of the weak base ammonia with the strong acid HCl:

[latex]\text{NH}_3(aq)\;+\;\text{HCl}(aq)\;{\longrightarrow}\;\text{NH}_4\text{Cl}(aq)[/latex]

A solution of this table salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a potent acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acrid of ammonia, reacts with water and increases the hydronium ion concentration:

[latex]\text{NH}_4^{\;\;+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{NH}_3(aq)[/latex]

The equilibrium equation for this reaction is just the ionization constant. K _a, for the acid [latex]\text{NH}_4^{\;\;+}[/latex]:

[latex]\frac{[\text{H}_3\text{O}^{+}][\text{NH}_3]}{[\text{NH}_4^{\;\;+}]} = K_{\text{a}}[/latex]

We will not find a value of M _a for the ammonium ion in Appendix H. However, it is not difficult to decide K _a for [latex]\text{NH}_4^{\;\;+}[/latex] from the value of the ionization constant of water, Grand _w, and K _b, the ionization constant of its conjugate base, NH₃, using the following relationship:

[latex]K_{\text{westward}} = K_{\text{a}}\;\times\;K_{\text{b}}[/latex]

This relation holds for whatever base and its conjugate acid or for any acrid and its conjugate base.

Example 1

The pH of a Solution of a Salt of a Weak Base and a Potent Acid
Aniline is an amine that is used to manufacture dyes. It is isolated every bit aniline hydrochloride, [latex][\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}]\text{Cl}[/latex], a common salt prepared by the reaction of the weak base of operations aniline and hydrochloric acid. What is the pH of a 0.233 Yard solution of aniline hydrochloride?

[latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\leftrightharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{C}_6\text{H}_5\text{NH}_2(aq)[/latex]

Solution
The new footstep in this example is to determine K _a for the [latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}[/latex] ion. The [latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}[/latex] ion is the conjugate acrid of a weak base. The value of M _a for this acrid is not listed in Appendix H, just we can determine it from the value of K _b for aniline, C_viH₅NH₂, which is given equally 4.3 × 10⁻¹⁰ (Table 3 in Chapter 14.3 Relative Strengths of Acids and Bases and Appendix I):

[latex]K_{\text{a}}\;(\text{for\;C}_6\text{H}_5\text{NH}_3^{\;\;+})\;\times\;K_{\text{b}}\;(\text{for\;C}_6\text{H}_5\text{NH}_2) = K_{\text{w}} = 1.0\;\times\;10^{-fourteen}[/latex]

[latex]K_{\text{a}}\;(\text{for\;C}_6\text{H}_5\text{NH}_3^{\;\;+}) = \frac{K_{\text{w}}}{K_{\text{b}}\;(\text{for\;C}_6\text{H}_5\text{NH}_2)} = \frac{1.0\;\times\;10^{-14}}{4.3\;\times\;x^{-10}} = 2.iii\;\times\;10^{-5}[/latex]

At present we have the ionization constant and the initial concentration of the weak acid, the data necessary to determine the equilibrium concentration of H_threeO⁺, and the pH:

With these steps we find [H₃O⁺] = 2.iii × 10^−three Thousand and pH = ii.64

Check Your Learning
(a) Do the calculations and show that the hydronium ion concentration for a 0.233-Thou solution of [latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}[/latex] is 2.3 × 10⁻³ and the pH is 2.64.

(b) What is the hydronium ion concentration in a 0.100-Thou solution of ammonium nitrate, NH₄NO₃, a common salt composed of the ions [latex]\text{NH}_4^{\;\;+}[/latex] and [latex]\text{NO}_3^{\;\;-}[/latex]. Use the information in Table 3 in Affiliate 14.three Relative Strengths of Acids and Bases to determine Yard _b for the ammonium ion. Which is the stronger acid [latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}[/latex] or [latex]\text{NH}_4^{\;\;+}[/latex]?

Reply:

(a) [latex]K_{\text{a}}\;(\text{for\;NH}_4^{\;\;+}) = v.6\;\times\;10^{-ten}[/latex], [H_iiiO⁺] = 7.v × 10⁻⁶ M; (b) [latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}[/latex] is the stronger acrid.

Salts of Weak Acids and Potent Bases

When nosotros neutralize a weak acid with a strong base of operations, nosotros get a table salt that contains the conjugate base of the weak acrid. This cohabit base is normally a weak base. For example, sodium acetate, NaCH₃CO_two, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:

[latex]\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{NaOH}(aq)\;{\longrightarrow}\;\text{NaCH}_3\text{CO}_2(aq)\;+\;\text{H}_2\text{O}(aq)[/latex]

A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acrid of a strong base, has no effect on the acidity of the solution. Even so, the acetate ion, the cohabit base of acetic acid, reacts with water and increases the concentration of hydroxide ion:

[latex]\text{CH}_3\text{CO}_2^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(50)\;{\rightleftharpoons}\;\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{OH}^{-}(aq)[/latex]

The equilibrium equation for this reaction is the ionization abiding, Grand _b, for the base [latex]\text{CH}_3\text{CO}_2^{\;\;-}[/latex]. The value of K _b can be calculated from the value of the ionization constant of water, K _{due west}, and K _a, the ionization constant of the cohabit acid of the anion using the equation:

[latex]K_{\text{w}} = K_{\text{a}}\;\times\;K_{\text{b}}[/latex]

For the acetate ion and its cohabit acrid we have:

[latex]K_{\text{b}}\;(\text{for\;CH}_3\text{CO}_2^{\;\;-}) = \frac{K_{\text{west}}}{K_{\text{a}}\;(\text{for\;CH}_3\text{CO}_2\text{H})} = \frac{1.0\;\times\;ten^{-xiv}}{1.8\;\times\;10^{-5}} = v.6\;\times\;10^{-10}[/latex]

Some handbooks do not report values of K _b. They only report ionization constants for acids. If we want to determine a K _b value using one of these handbooks, nosotros must look up the value of K _a for the conjugate acid and catechumen it to a K _b value.

Example 2

Equilibrium in a Solution of a Salt of a Weak Acrid and a Strong Base
Make up one's mind the acetic acid concentration in a solution with [latex][\text{CH}_3\text{CO}_2^{\;\;-}] = 0.050\;One thousand[/latex] and [OH⁻] = 2.5 × ten^−six Grand at equilibrium. The reaction is:

[latex]\text{CH}_3\text{CO}_2^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{OH}^{-}(aq)[/latex]

Solution
We are given ii of three equilibrium concentrations and asked to observe the missing concentration. If we tin can find the equilibrium constant for the reaction, the process is straightforward.

The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We decide K _b equally follows:

[latex]K_{\text{b}}\;(\text{for\;CH}_3\text{CO}_2^{\;\;-}) = \frac{K_{\text{w}}}{K_{\text{a}}\;(\text{for\;CH}_3\text{CO}_2\text{H})} = \frac{1.0\;\times\;10^{-fourteen}}{i.8\;\times\;10^{-five}} = 5.6\;\times\;10^{-10}[/latex]

Now detect the missing concentration:

[latex]K_{\text{b}} = \frac{[\text{CH}_3\text{CO}_2\text{H}][\text{OH}^{-}]}{[\text{CH}_3\text{CO}_2^{\;\;-}]} = 5.six\;\times\;10^{-x}[/latex]

[latex]= \frac{[\text{CH}_3\text{CO}_2\text{H}](ii.5\;\times\;ten^{-6})}{(0.050)} = v.6\;\times\;10^{-10}[/latex]

Solving this equation we become [CH₃CO₂H] = 1.1 × ten⁻⁵ Yard.

Cheque Your Learning
What is the pH of a 0.083-G solution of CN⁻? Utilise iv.nine × 10^−ten as K _a for HCN. Hint: Nosotros volition probably need to convert pOH to pH or find [H₃O⁺] using [OH⁻] in the final stages of this problem.

Equilibrium in a Solution of a Salt of a Weak Acrid and a Weak Base of operations

In a solution of a table salt formed by the reaction of a weak acrid and a weak base, to predict the pH, we must know both the Thousand _a of the weak acid and the K _b of the weak base of operations. If K _a > K _b, the solution is acidic, and if Chiliad _b > Chiliad _a, the solution is basic.

Case 3

Determining the Acidic or Basic Nature of Salts
Make up one's mind whether aqueous solutions of the post-obit salts are acidic, basic, or neutral:

(a) KBr

(b) NaHCO_iii

(c) NH₄Cl

(d) Na₂HPO₄

(e) NH_fourF

Solution
Consider each of the ions separately in terms of its issue on the pH of the solution, as shown here:

(a) The Grand⁺ cation and the Br⁻ anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acrid (HBr), respectively. The solution is neutral.

(b) The Na⁺ cation is a spectator, and volition non affect the pH of the solution; while the [latex]\text{HCO}_3^{\;\;-}[/latex] anion is amphiprotic, information technology could either behave every bit an acid or a base. The K _a of [latex]\text{HCO}_3^{\;\;-}[/latex] is 4.7 × ten^−eleven, so the G _b of its conjugate base is [latex]\frac{1.0\;\times\;10^{-fourteen}}{4.3\;\times\;10^{-seven}} = 2.3\;\times\;10^{-8}[/latex].

Since M _b >> K _a, the solution is basic.

(c) The [latex]\text{NH}_4^{\;\;+}[/latex] ion is acidic and the Cl⁻ ion is a spectator. The solution will be acidic.

(d) The Na⁺ ion is a spectator, while the [latex]\text{HPO}_4^{\;\;2-}[/latex] ion is amphiprotic, with a K _a of iv.2 × 10^−xiii so that the M _b of its conjugate base is [latex]\frac{ane.0\;\times\;ten^{-14}}{6.2\;\times\;x^{-eight}} = 1.half dozen\;\times\;ten^{-7}[/latex]. Because K _b >> M _a, the solution is basic.

(e) The [latex]\text{NH}_4^{\;\;+}[/latex] ion is listed as being acidic, and the F⁻ ion is listed every bit a base, so we must straight compare the One thousand _a and the K _b of the ii ions. K _a of [latex]\text{NH}_4^{\;\;+}[/latex] is five.six × ten⁻¹⁰, which seems very small, yet the K _b of F⁻ is 1.four × ten⁻¹¹, so the solution is acidic, since K _a > K _b.

Check Your Learning
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

(a) K₂CO₃

(b) CaCl_two

(c) KH₂PO₄

(d) (NH₄)₂CO₃

(e) AlBr₃

Answer:

(a) bones; (b) neutral; (c) basic; (d) basic; (e) acidic

The Ionization of Hydrated Metal Ions

If we measure out the pH of the solutions of a variety of metal ions we will find that these ions deed as weak acids when in solution. The aluminum ion is an instance. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, [latex]\text{Al(H}_2\text{O})_6^{\;\;3+}[/latex], dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the then-called start solvation crush), even though it does interact with the other h2o molecules surrounding this [latex]\text{Al(H}_2\text{O})_6^{\;\;3+}[/latex] cluster equally well:

[latex]\text{Al(NO}_3)_3(s)\;+\;half dozen\text{H}_2\text{O}(50)\;{\longrightarrow}\;\text{Al(H}_2\text{O})_6^{\;\;three+}(aq)\;+\;iii\text{NO}_3^{\;\;-}(aq)[/latex]

We frequently see the formula of this ion but equally "Al^three+(aq)", without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, H₃O⁺ to H⁺. However, in this instance, the hydrated aluminum ion is a weak acid (Figure 2) and donates a proton to a water molecule. Thus, the hydration becomes important and we may apply formulas that show the extent of hydration:

[latex]\text{Al(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_5(\text{OH})^{2+}(aq)\;\;\;\;\;\;\;K_{\text{a}} = 1.four\;\times\;x^{-5}[/latex]

As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown past:

[latex]\begin{array}{r @{{}\rightleftharpoons{}} l} \text{Al(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_5(\text{OH})^{2+}(aq) \\[0.5em] \text{Al(H}_2\text{O})_5(\text{OH})^{ii+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_4(\text{OH})_2^{\;\;+}(aq) \\[0.5em] \text{Al(H}_2\text{O})_4(\text{OH})_2^{\;\;+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_3(\text{OH})_3(aq) \end{array}[/latex]

Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the left side of the equilibrium expressions and every bit bases when they appear on the right side.

Figure 2. When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

Still, the ionization of a cation carrying more than one charge is usually non extensive across the first phase. Additional examples of the first stage in the ionization of hydrated metal ions are:

[latex]\begin{assortment}{r @{{}\rightleftharpoons{}} ll} \text{Atomic number 26(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Fe(H}_2\text{O})_5(\text{OH})^{2+}(aq) & K_{\text{a}} = 2.74 \\[0.5em] \text{Cu(H}_2\text{O})_6^{\;\;2+}(aq)\;+\;\text{H}_2\text{O}(fifty) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Cu(H}_2\text{O})_5(\text{OH})^{+}(aq) & K_{\text{a}} = {\sim}vi.3 \\[0.5em] \text{Zn(H}_2\text{O})_4^{\;\;2+}(aq)\;+\;\text{H}_2\text{O}(fifty) & \text{H}_3\text{O}^{+}(aq)\;+\;\text{Zn(H}_2\text{O})_3(\text{OH})^{+}(aq) & K_{\text{a}} = 9.vi \stop{array}[/latex]

Instance 4

Hydrolysis of [Al(H₂O)₆]ⁱⁱⁱ⁺
Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [latex][\text{Al(H}_2\text{O})_6]^{3+}[/latex] in solution.

Solution
In spite of the unusual appearance of the acid, this is a typical acrid ionization problem.

1. Determine the direction of change. The equation for the reaction and M _a are:

   [latex]\text{Al(H}_2\text{O})_6^{\;\;3+}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{Al(H}_2\text{O})_5(\text{OH})^{two+}(aq)\;\;\;\;\;\;\;K_{\text{a}} = 1.4\;\times\;10^{-5}[/latex]

   The reaction shifts to the correct to achieve equilibrium.

2. Determine x and equilibrium concentrations. Apply the tabular array:
3. Solve for x and the equilibrium concentrations. Substituting the expressions for the equilibrium concentrations into the equation for the ionization abiding yields:

   [latex]K_{\text{a}} = \frac{[\text{H}_3\text{O}^{+}][\text{Al(H}_2\text{O})_5(\text{OH})^{2+}]}{[\text{Al(H}_2\text{O})_6^{\;\;3+}]}[/latex]

   [latex]= \frac{(x)(x)}{0.10\;-\;x} = 1.4\;\times\;10^{-5}[/latex]

   Solving this equation gives:

   [latex]x = 1.2\;\times\;ten^{-3}\;M[/latex]

   From this we find:

   [latex][\text{H}_3\text{O}^{+}] = 0\;+\;x = 1.ii\;\times\;10^{-3}\;M[/latex]

   [latex]\text{pH} = -\text{log[H}_3\text{O}^{+}] = 2.92\;(\text{an\;acidic\;solution})[/latex]

4. Cheque the work. The arithmetic checks; when 1.2 × 10⁻³ M is substituted for ten, the outcome = K _a.

Check Your Learning
What is [latex][\text{Al(H}_2\text{O})_5(\text{OH})^{ii+}][/latex] in a 0.15-Chiliad solution of Al(NO₃)₃ that contains enough of the strong acid HNO₃ to bring [H_iiiO⁺] to 0.10 M?

The constants for the different stages of ionization are not known for many metallic ions, so we cannot calculate the extent of their ionization. Nevertheless, practically all hydrated metal ions other than those of the brine metals ionize to requite acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.

Primal Concepts and Summary

The characteristic properties of aqueous solutions of Brønsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Brønsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may brand the product solutions slightly acidic or slightly basic.

Solutions that comprise salts or hydrated metal ions accept a pH that is determined by the extent of the hydrolysis of the ions in the solution. The pH of the solutions may exist calculated using familiar equilibrium techniques, or information technology may be qualitatively adamant to be acidic, basic, or neutral depending on the relative Thousand _a and K _b of the ions involved.

Chemistry End of Chapter Exercises

1. Determine whether aqueous solutions of the post-obit salts are acidic, basic, or neutral:

   (a) Al(NO_iii)₃

   (b) RbI

   (c) KHCO₂

   (d) CH₃NH₃Br

2. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

   (a) FeCl_three

   (b) Thou₂CO₃

   (c) NH_ivBr

   (d) KClO₄

3. Novocaine, C₁₃H₂₁O_iiN_iiCl, is the table salt of the base procaine and muriatic acid. The ionization abiding for procaine is seven × ten^−six. Is a solution of novocaine acidic or basic? What are [H₃O⁺], [OH⁻], and pH of a 2.0% solution past mass of novocaine, assuming that the density of the solution is one.0 k/mL.

Solutions

Answers to Chemistry End of Chapter Exercises
2. (a) acidic; (b) basic; (c) acidic; (d) neutral

deanabightly.blogspot.com

Source: https://opentextbc.ca/chemistry/chapter/14-4-hydrolysis-of-salt-solutions/

Share this post