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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

 p^H = -\log (2\times 10^{-13})=12.69


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