Concepts: buffers, salt hydrolysis and solubility product

Equilibrium24 min readLesson 5 of 8

5.1 Buffer solutions A buffer solution resists a change in pH when a small amount of acid or base is added, or when the solution is diluted. An acidic buffer contains a weak acid and a salt of its conjugate base, for example acetic acid and sodium acetate. A basic buffer contains a weak base and its salt, for example ammonia and ammonium chloride. The weak acid supplies a reservoir of undissociated molecules that can react with added base, and the conjugate base supplies a reservoir that can react with added acid. Adding hydrogen ions converts some acetate into acetic acid, and adding hydroxide ions converts some acetic acid into acetate. The ratio of the two components changes only slightly, so the hydrogen ion concentration changes only slightly. The pH of a buffer is calculated from the dissociation constant of the weak acid and the ratio of the salt concentration to the acid concentration. The same relation is often quoted with the logarithm of the ratio of the conjugate base to the acid.

p H = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}

Henderson-Hasselbalch equation for an acidic buffer

When the salt and acid concentrations are equal, the logarithm is zero and the pH equals the pKa. A buffer works best when the ratio lies between about 0.1 and 10, because outside that range one component dominates and the buffer capacity becomes small. Buffer capacity also grows with the total concentration of the two components. A basic buffer follows the same logic with the base dissociation constant. For ammonia and ammonium chloride, the pOH is the pKb plus the logarithm of the ratio of the salt concentration to the base concentration, and the pH follows from pOH.

pO H = p K_{b} + lo g \frac{[ salt ]}{[ base ]} p H = 14 - pO H

Buffer of a weak base and its salt

Blood is a familiar buffered system. The carbonic acid and bicarbonate pair keeps the pH of blood close to 7.4, and a deviation of a few tenths of a unit can be dangerous. The dihydrogen phosphate and hydrogen phosphate pair buffers the inside of cells. 5.2 Salt hydrolysis A salt formed from a strong acid and a strong base gives a neutral solution, because neither ion hydrolyses. Sodium chloride is the standard example. The other three combinations give acidic or basic solutions because one of the ions reacts with water. A salt of a weak acid and a strong base hydrolyses to give a basic solution. Sodium acetate is the example: the acetate ion accepts a proton from water, producing hydroxide ions, so the solution is alkaline. The hydrolysis constant is the ionic product divided by the acid dissociation constant of the weak acid.

K_{h} = \frac{K _{w}}{K _{a}} p H = \frac{1}{2} (14 + p K_{a} + lo g c)

Hydrolysis constant and pH for a weak acid strong base salt

A salt of a strong acid and a weak base gives an acidic solution. Ammonium chloride is the standard example: the ammonium ion donates a proton to water and the solution becomes acidic. The hydrolysis constant is Kw divided by Kb.

K_{h} = \frac{K _{w}}{K _{b}} p H = \frac{1}{2} (14 - p K_{b} - lo g c)

Hydrolysis constant and pH for a strong acid weak base salt

A salt of a weak acid and a weak base hydrolyses on both sides, so the pH depends on the relative strengths of the two ions and not on the concentration. The pH is approximately half the sum of the two pK values. The buffer capacity is worth distinguishing from the buffer pH. The pH is set by the pKa and the ratio, while the capacity is set by the absolute concentrations, because a more concentrated buffer can absorb more added acid or base before the ratio changes appreciably. 5.3 Solubility product A sparingly soluble salt in contact with its saturated solution establishes an equilibrium between the solid and its ions. The equilibrium constant for this process is the solubility product, written from the ion concentrations raised to the powers of their coefficients. The solid does not appear in the expression.

K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}

Solubility product of a salt A x B y

For a salt that dissociates into one cation and one anion, the solubility product is the square of the solubility. For a salt that gives two ions of one kind and one of the other, the solubility product is four times the cube of the solubility. Knowing Ksp allows the solubility and the molar concentrations to be calculated, and knowing the solubility allows Ksp to be found.

A B : K_{s p} = s^{2} A B_{2} or A_{2} B : K_{s p} = 4 s^{3} A_{3} B_{2} or A_{2} B_{3} : K_{s p} = 108 s^{5}

Solubility products from solubility s for common salt types

The common ion effect follows at once. Adding a salt that supplies one of the ions shifts the solubility equilibrium to the left, so the salt becomes less soluble. Adding silver nitrate to a saturated silver chloride solution lowers the chloride concentration because the silver ion is now in excess and the equilibrium moves to consume it. This is why a precipitate forms when a soluble salt of the common ion is added. The reaction quotient for a dissolving salt is the ion product compared with Ksp. If the ion product is smaller than Ksp the solution is unsaturated and no precipitate forms, if it equals Ksp the solution is exactly saturated, and if it exceeds Ksp the salt precipitates until the product falls back to Ksp. This comparison is the basis of qualitative analysis and of the selective precipitation used to separate ions.

Exam

In buffer and hydrolysis problems, watch the powers and roots in the pH formulas. A buffer uses the salt-to-acid ratio directly, while hydrolysis of a weak acid strong base salt contains the square root of c and half of the pKa. Student errors here are usually algebra, not chemistry.

in hinglish

Buffer solution pH ke change ko resist karta hai: acidic buffer mein weak acid + uska salt (jaise CH3COOH + CH3COONa), basic buffer mein weak base + uska salt (NH3 + NH4Cl). pH nikaalne ke liye Henderson equation : pH = pKa + log([salt]/[acid]); jab dono equal ho to pH = pKa. Salt hydrolysis char cases: strong acid + strong base \to neutral; weak acid + strong base \to basic (Kh = Kw/Ka); strong acid + weak base \to acidic (Kh = Kw/Kb); weak acid + weak base \to pH dono ki relative strength par. Solubility product Ksp sirf saturated solution ke ions se banta hai, solid nahi aata. AB ke liye Ksp = s², AB2 ya A2B ke liye 4s³, A3B2 ke liye 108s⁵. Common ion effect se solubility ghatti hai kyunki equilibrium peeche shift hota hai. Q vs Ksp se decide karo: Q < Ksp matlab unsaturated, Q > Ksp matlab precipitate.