Answer:
pH = 12.43
Explanation:
<em>...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</em>.
To solve this question we need to know that hidrazoic acid reacts with KOH as follows:
HN3 + KOH → KN3 + H2O
<em>Moles KOH:</em>
0.5716L * (0.2900mol /L) =0.1658 moles of KOH
<em>Moles HN3:</em>
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
