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# If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions.

7.57     If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions.What is its pH?

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We have 0.562 g of potassium hydroxide ($KOH$). On dissolving in water gives 200 mL of solution.
Therefore, concentration of $[KOH(aq)]$ = $\frac{0.561\times 1000}{200}g/L$

= 2.805 g/L

$=2.805 \times \frac{1}{56.11} M=0.05M$
$KOH(aq) \rightarrow K^+(aq) + OH^-(aq)$
It is a strong base. So, that it goes complete dissociation.

$[OH^-] = 0.05M = [K^+]$

It is known that,

$\\K_w = [H^+][OH^-]\\ {[H^+]}=\frac{K_w}{[OH^-]}$
$=\frac{10^{-14}}{0.05} = 2\times 10^{-13}M$

Therefore,
$p^H = -\log (2\times 10^{-13})=12.69$

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