MCQ
Given that the dissociation constant for ${H_2}O$ is ${K_w} = 1 \times {10^{ - 14}}$ $mol{e^2}\,\,litr{e^{ - 2}},$ what is the $HCl$ of a $0.001\,\,molar\,\,KOH$ solution
- A${10^{ - 11}}$
- B$3$
- C$14$
- ✓$11$
✓
Answer
Correct option: D.
$11$
(d) $0.001\, M KOH$ solution
$[O{H^ - }] = 0.001\,M = 1 \times {10^{ - 3}}\,M$
$[{H^ + }] \times [O{H^ - }] = 1 \times {10^{ - 14}}$
$[{H^ + }] = \frac{{1 \times {{10}^{ - 14}}}}{{[O{H^ - }]}}$
$[{H^ + }] = \frac{{1 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 3}}}}$$ = 1 \times {10^{ - 14}} \times {10^{ + 3}}$
$[{H^ + }] = {10^{ - 11}}\,M$
$pH = 11$
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