Question

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200.00mL of 2.00 M solution of a weak acid is mixed with 300.00 mL of a 2.00 M solution containing its conjugate base. 50.000 g of the slightly soluble salt silver acetate (AgC2H3O2) is added to this buffer solution. After thorough mixing, the solution is assumed to be saturated. To determine the amount of the solid which remains undissolved, the solution is filtered, and the residue is collected and dried; the mass of the undissolved solid is 30.369 g.

a)Based on this information, Calculate Ksp' for silver acetate in this buffer. (*Note Ksp PRIME is needed, not just regular Ksp)

b)Using your value for Ksp', determine the pH of the buffer solution.

c)Based on the pH of the buffer solution, identify the weak acid used to produce it.

Answer 1

A general approach of this question:

The buffer will limit the available concentration of CH₃COO^- (the dissociation of CH3COOH is controlled by [H⁺] in the buffer). If CH₃COO^- is consumed (neutralized) by the buffer, you may expect a higher apparent solubilty of CH₃COOAg than the regular one (le Chatelier’s principle is also relevant here).

The buffered concentration of [H⁺] regulates both dissociation equilibria (of CH3COOH and of the unknown weak acid AH)

Answer:

a)

Molar mass of CH₃COOAg is 166.91 g/mol.

50.000 – 30.369 = 19.631 g CH₃COOAg are dissolved, so

19.631 g/166.91 g/mol = 0.11761 mol CH₃COOAg are dissolved in (300+200)= 500 mL solution.

Then: [CH₃COO^- ] = 0.11761 mol/0.5L = 0.235 M = S’

This value is the apparent/conditioned molar solubility S’ of CH₃COO^-.

From the reaction

CH3COOAg   < == > CH₃COO^- + Ag⁺

for a regular situation:

[CH₃COO^- ]=[Ag⁺] = S

K_sp = [CH₃COO^- ][Ag⁺] = S² = 1.94 x 10^-3 as a regular value (from Tables).

By analogy, here:

K_sp’ = (S’)² = [CH₃COO^- ]² = 0.235² = 5.53x10^-2        (K_sp’ > K_sp as expected)

b)

For CH3COOH pKa = 4.75 and Ka = 1.76x10^-5

[H⁼] = Ka/[CH₃COO^-] = Ka/( K_sp’)^1/2 = 1.76x10^-5 / 0.235 = 7.49x10^-5 M

pH = -log7.49x10^-5 M = 5-0.875= 4.125

c)

For the unknown weak acid AH :

The buffer contains:

2.00 M x 0.300 L = 0.6 mol A^-

2.00 M x 0.200 L = 0.4 mol AH

[A-]/[AH] = 0.6/0.4 = 1.500

log [A-]/[AH] = log 1.5 = 0.1761

From the general formula of pH of a buffer mixture is

pH = pK_a,HX + log([base]/[acid])

pK_a,HX = pH - log([base]/[acid])

            = 4.125 - 0.176 = 3.948

The “identity” of the unknown acid is pKa = 3.95