Question

29. What is the pH of an aqueous solution made by combining 40.03 mL of a 0.4322 M sodium fluoride with 44.86 mL of a 0.3356 M solution of hydrofluoric acid to which 3.619 mL of a 0.0543 M solution of HCl was added?

Answer 1

millimoles of NaF = 40.03 x 0.4322 = 17.301

millimoles of HF = 44.86 x 0.3356 = 15.055

millimoles of HCl added = 3.619 x 0.0543 = 0.196

after HCl added

millimoles of NaF = 17.301 - 0.196 = 17.105

millimoles of HF = 15.055 + 0.196 = 15.251

total volume = 40.03 + 44.86 + 3.619 = 88.509 mL

[NaF] = 17.105 / 88.509 = 0.1932 M

[HF] = 15.251 / 88.509 = 0.1723 M

pKa of HF = 3.17 standard value

mixture of HF and NaF act as acidic buffer

pH = pKa + log [NaF] / [HF]

pH = 3.17 + log [0.1932] / [0.1723]

pH = 3.22