Question

(d) Calculate the pH of the solution that results from mixing 1.53 mL of 0.179 M HCl with 107.0 mL of 0.0160 M NaOH.

(e) Calculate the pH of the solution that results from mixing 13.04 mL of 0.111 M HCl with 81.0 mL of a 0.0151 M solution of a weak base whose pKb = 5.81.

(f) Calculate the pH of the solution that results from mixing 3.63 mL of 0.111 M HCl with 81.0 mL of a 0.0151 M solution of a weak base whose pKb = 5.81.

Answer 1

d)

mol of HCl = MV = 1.53*0.179 = 0.27387

mol of base = MV = 107*0.016 = 1.712

mol of base left = 1.712-0.27387 = 1.43813

[OH-] = mol/V = 1.43813/(1.53+107) = 0.01325

pOH = -log(0.01325 =

pH = 14-pOH = 14-1.877

pH = 12.12

e)=

mmol of acid = MV = 13.04*0.111 = 1.44744

mmol of base = M V= 81*0.0151 = 1.2231

after reaction

mmol of acid left = 1.44744-1.2231 = 0.22434

[H+] = 0.22434 / (13.04+81) = 0.0023855

pH = -log(0.0023855) = 2.62

f)

mmol of base = M V= 81*0.0151 = 1.2231

mmol of acid = MV = 0.111*3.63 = 0.40293

mmol of bae left = 1.2231-0.40293 = 0.82017

mmol of conjguate formed = 0.40293

pH = pKa + log(B/BH+)

pH = (14-5.81) + log(0.40293/0.82017)

pH = 7.881