Answer:
pH = 12.43
Explanation:
...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
Moles KOH:
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
Moles HN3:
0.2127L * (0.6800mol/L) = 0.1446 moles HN3
As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:
0.1658 moles - 0.1446 moles =
0.0212 mol KOH
In 212.7mL + 571.6mL = 784.3mL = 0.7843L
The molarity of KOH = [OH-] is:
0.0212 mol KOH / 0.7843L = 0.027M = [OH-]
The pOH is defined as -log [OH-]
pOH = -log 0.027M
pOH = 1.57
pH = 14 - pOH
pH = 12.43
