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What is the $[OH^-]$ in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M $Ba(OH)_2$?
A
0.12 M
B
0.10 M
C
0.40 M
D
0.0050 M
Detailed Solution
Millimoles of $H^+$ from HCl = $20.0\times0.050 = 1.0$ mmol
Each $Ba(OH)_2$ gives 2 $OH^-$: millimoles of $OH^-$ = $30.0\times0.10\times2 = 6.0$ mmol
$H^+ + OH^- \rightarrow H_2O$: 1.0 mmol of $H^+$ neutralises 1.0 mmol of $OH^-$
$OH^-$ left = 6.0 − 1.0 = 5.0 mmol
Total volume = 20.0 + 30.0 = 50.0 mL
$[OH^-] = \frac{5.0}{50.0} = 0.10$ M
Each $Ba(OH)_2$ gives 2 $OH^-$: millimoles of $OH^-$ = $30.0\times0.10\times2 = 6.0$ mmol
$H^+ + OH^- \rightarrow H_2O$: 1.0 mmol of $H^+$ neutralises 1.0 mmol of $OH^-$
$OH^-$ left = 6.0 − 1.0 = 5.0 mmol
Total volume = 20.0 + 30.0 = 50.0 mL
$[OH^-] = \frac{5.0}{50.0} = 0.10$ M