Question

I have a mixture consisting of 0.180 M nitrous acid, and 0.150 M sodium nitrite. NOTE: pKa (nitrous acid) = 3.14. 7.

First, I add 2.00 mL of 0.500 M HCl to 20.00 mL of the original (sodium nitrite / nitrous acid) mixture.

Secondly, I add 2.00 mL of 0.500 M HCl (aq) to 20.00 mL of H2O.

Thirdly, I add 2.00 mL of 0.500 M NaOH (aq) to 20.00 mL of the original (sodium nitrite / nitrous acid) mixture.

Finally, I add 2.00 mL of 0.500 M NaOH (aq) to 20.00 mL of H2O.

Assuming these volumes are additive, what is the new pH of the resulting mixture in this case?

Answer 1

pH = pKa + log (salt) - (acid) / (acid) + (acid)

pH = 3.14 + log ((0.15 * 20 - 0.5 * 2) / (0.18*20 + 0.5 *2 ))

pH = 2.778

concentration of HCl = 0.5 * 2 / (20 +2) = 0.0454

pH = 3.14 + log ((0.15 * 20 - 0.0454 ) / (0.18*20 + 0.0454 ))

pH = 3.0487

pH = pKa + log (salt) + (base) / (acid) - (base)

pH = 3.14 + log ((0.15 * 20 + 0.5 * 2) / (0.18*20 - 0.5 *2 ))

pH = 3.32

concentration of NaOH = 0.5 * 2 / (20 +2) = 0.0454

pH = 3.14 + log ((0.15 * 20 + 0.0454 ) / (0.18*20 - 0.0454 ))

pH = 3.072