Calculate the pH at the equivalence point for the titration of 0.22 M HCN with 0.22 M NaOH. (K_a = 4.9 * 10^-10 for HCN)

Question

Quizplus · Accepted Answer

The equation for the reaction between HCN and NaOH is:
HCN + NaOH → NaCN + H2O

For every 1 mole of NaOH added, 1 mole of HCN will react. At the equivalence point, all the HCN has reacted with NaOH and been converted to NaCN. NaCN is the salt of a weak acid (HCN) and a strong base (NaOH), making the solution basic.

To determine the pH at the equivalence point, we can use the equation for the ionization constant of water:
Kw = [H+][OH-] = 1.0x10^-14

At the equivalence point, [HCN] = [NaOH] = 0.22 M. Since NaOH is a strong base, it completely dissociates in water to form OH- ions. Therefore, [OH-] = 0.22 M.

To find [H+], we can use the expression for the ionization constant (Ka) of HCN:
Ka = [H+][CN-]/[HCN]

At the equivalence point, [HCN] = 0. By stoichiometry, all of the HCN has reacted with NaOH to form NaCN. Therefore, [CN-] = 0.22 M.

Substituting into the expression for Ka, we get:
4.9x10^-10 = [H+][0.22]/0

Solving for [H+], we get:
[H+] = 0

This means that the solution is completely basic at the equivalence point and the pH is determined by the concentration of the hydroxide ions. Using the expression for pH, we get:

pH = -log[OH-] = -log(0.22) = 1.65

Finally, we need to add 14 to the pH to get the pOH and then subtract the pOH from 14 to get the pH:
pOH = 14 - pH = 14 - 1.65 = 12.35
pH = 14 - pOH = 14 - 12.35 = 1.65

Therefore, the pH at the equivalence point is 11.18 (choice D).

Calculate the PH at the Equivalence Point for the Titration