Question

1. A buffer that contains 0.37 M of an acid, HA and 0.48 M of its conjugate base A^-, has a pH of 3.76. What is the pH after 0.02 mol of NaOH are added to 0.71 L of the solution?

2. Calculate the pH during the titration of 30 mL of 0.25 M HNO₃(aq) with 0.18 M NaOH after 18 mL of the base have been added.

Answer 1

1) first calculate the pKa by using given values

pH = pKa + log [A^-] / [HA]

3.76 = pKa + log [0.48] / [0.37]

3.76 = pKa + 0.11

pKa = 3.65

now after 0.02 / 0.71 = 0.028 M NaOH added

[A^-] = 0.48 + 0.028 = 0.508 M

[HA] = 0.37 + 0.028 = 0.398 M

pH = pKa + log [A^-] / [HA]

pH = 3.65 + log [0.508] / [0.398]

pH = 3.75

2) HNO3 + NaOH ------------> NaNO3 + H2O

millimoles of HNO3 = 30 x 0.25 = 7.5

millimoles of NaOH added = 18 x 0.18 = 3.24

7.5 - 3.24 = 4.26 millimoles HNO3 left

[HNO3] = 4.26 / 18 + 30 = 0.089 M

pH = - log [H⁺]

pH = - log [0.089]

pH = 1.05