Question

1. Consider a reaction mixture containing 100.0 mL of 0.112 M borate buffer at pH = pKa = 9.24. At pH = pKa, we know that [H3BO3] = [H2BO3−] = 0.0560 M. Suppose that a chemical reaction whose pH we wish to control will be generating acid. To avoid changing the pH very much we do not want to generate more acid than would use up half of the [H2BO3−].

a)How many moles of acid could be generated without using up more than half of the [H2BO3−]?

b) What would be the pH?

2. Calculate the fraction of association (α) for the following concentrations of sodium butanoate. (Assume K_a = 1.52 ✕ 10⁻⁵ and K_w = 1.01 ✕ 10⁻¹⁴. Enter unrounded values.)

a)1.04 ✕ 10⁻¹M

b) 1.04 ✕ 10⁻²M

c) 1.04 ✕ 10⁻¹²M

Answer 1

Q.1: (a) Given the volume of the reaction mixture, V = 100.0 mL = 100.0 mL x (1L / 1000 mL) = 0.100 L

Concentration of H2BO3-, [H2BO3-] = 0.0560 M

Hecne moles of moles of H2BO3- in the mixture = MxV = 0.0560 M x 0.100 L = 0.00560 mol

moles of moles of H3BO3 in the mixture = MxV = 0.0560 M x 0.100 L = 0.00560 mol

We need to retain half of the moles of H2BO3-.

Hence moles of H2BO3- reacts with the acid formed = 0.00560 mol / 2 = 0.00280 mol

Hence moles of acid that could be generated = 0.00280 mol (answer)

(b) 0.00280 mol of the acid formed will react with 0.00280 mol of H2BO3- to form 0.00280 mol of H3BO3

-----------------H2BO3- + H+ ------> H3BO3

Init. moles:0.00560, 0.00280, 0.00560

mol aft.rxn:0.00280, -- 0 -------- (0.00560+0.00280)

Hence after reaction, [H2BO3-] = 0.00280 mol / 0.1L = 0.0280 M

[H3BO3] = (0.00560+0.00280) mol / 0.1L = 0.0840 M

Now pH can be calculated from Hendersen equation

pH = pKa + log[H2BO3-] / [H3BO3] = 9.24 + log(0.0280M / 0.0840M) = 8.76 (answer)

Q.2(a) : The association reaction of sodium butanoate is

-------------CH3CH2CH2COO^-(aq) + H2O --- > CH3CH2CH2COOH + OH^-(aq), Kb = Kw / Ka = 1.01x10^-14 / 1.52x10^-5

Init.conc(M): 1.04x10^-1 ------------------------------- 0 -------------------------- 0

eqm.conc(M):(1.04x10^-1 - ) ------------------------, --------------------------

=> Kb = 1.01x10^-14 / 1.52x10^-5 = 6.645x10^-10 = [CH3CH2CH2COOH] x[OH^-(aq)] / [CH3CH2CH2COO^-(aq)]

=>  6.645x10^-10 = ² / (1.04x10^-1 - )

=> ² + 6.645x10^-10x - 6.911x10^-11 = 0

=>  = 8.31 x 10^-6 (answer)

Similarly we can calculate for b and c by changing the concentration