Calculate the pH of the buffer after the addition of 1.079 mL of 11.508 M HCl to 1.00 L of the buffer.

Question

Rank the following bases in order of increasing equivalence point pH if titrated with HCl.

Calculate the pH of a buffer containing 0.182  M HNO₂ and 0.170 M NaNO₂ at 25°C. The pK_a value for HNO₂ is 3.40.

Calculate the pH of the buffer after the addition of 1.079 mL of 11.508 M HCl to 1.00 L of the buffer.

Answer 1

The pH of the buffer is obtained using the Henderson-Hasselbalch equation and yields pH = 3.370. This value is slightly lower than the pKa as expected when the concentration of the acid (0.182) is larger than that of the salt (0.170).

The pH after the addition of 1.079 mL of strong acid is pH = 3.365 and this is obtained from a generalized version of the Henderson-Hasselbalch that accounts for the moles of acid and salt after the addition. As expected, this new pH is slightly lower and therefore more acidic, than the pH of the buffer before the addition of 1.079 mL of the acid.

I hope this helps.

Thank you.

Answer 2

Henderson Hasselbalch equation: pH = pKa + log [conj.base] / [acid]

pH = 3.40 + log [0.170] / [ 0.182]

pH = 3.40 + log 0.934

pH = 3.40 - 0.03

pH = 3.37

After adding HCl, the pH will decrease slightly because we are adding acid (H₊). The H⁺ will react with the NO₂^- to form HNO₂, as seen in the ICE table below. The mols of acid (H⁺) being added is

1.079 mls x 1 L / 1000 mls x 11.508 mol / L = 0.0124 moles.

NO₂^- + H⁺ ==> HNO₂

0.170......0.0124......0.182.........Initial

-0.0124...-0.0124....+0.0124....Change

0.1576.......0..........0.1944.......Equilibrium

Using the Henderson Hasselbalch equation again, we can find the new pH:

pH = pKa + log [conj.base] / [acid]

pH = 3.40 + log [0.1576 / 0.1944]

pH = 3.40 + log 0.811

pH = 3.40 - 0.09

pH = 3.31