Question

A)what is the pH of a solution made by mixing 465 ml of .10 M hydrochloric acid and 285 ml of .15 M sodium hydroxide?

B)What is the pH of a buffer solution prepared from 0.21 mol NH3 and .39 mol NH4NO3 dissolved in 1.00L of solution?

C) what is the pH in part B if >0.050 mol of Hcl is added? if 0.050 mol of NaOH is added?

Answer 1

A) moles of HCl = 0.1 M x 465 ml = 46.5 mmol

moles of NaOH = 0.15 M x 285 ml = 42.75 mmol

excess HCl = 3.75 mmol

Concentration of exces acid in solution = 3.75/(465 + 285) = 5 x 10^-3 M

pH = -log(5 x 10^-3) = 2.30

B) pH of buffer,

pH = pKa + log(base/acid)

[NH3] = 0.21 mol/1 L = 0.21 M

[NH4+] = 0.39 mol/1 L = 0.39 M

pKa = 14 - pKb = 14 - 4.75 = 9.25

So pH = 9.25 + log(0.21/0.39) = 8.98

C) When 0.05 mol HCl is adde

New [NH3] = (0.21 - 0.05)/1 L = 0.16 M

New [NH4+] = (0.39 + 0.05)/1 L = 0.44 M

New pH = 9.25 + log(0.16/0.44) = 8.81

When 0.05 mol of NaOh is added

New [NH3] = (0.21 + 0.05)/1 L = 0.26 M

New [NH4+] = (0.39 - 0.05)/1 L = 0.34 M

New pH = 9.25 + log(0.26/0.34) = 9.13