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?

Asked by Pragya Singh | 1 year ago |  82

##### Solution :-

[KOH(aq)

=  $$\dfrac{ 0.561}{(\dfrac{1}{5})}g/L$$

= 2.805 g/L

= 2.805 x ($$\dfrac{1}{56.11}$$)

= 0.05M

KOH(aq) → K+ (aq) + OH(aq)

[OH] = 0.05M = [K+] [H+][OH] = Kw

$$[H^{+}]$$ = $$\dfrac{ K_w}{[OH^{–}] [H^{+}]}$$

$$=\dfrac{10^{-14}}{0.05}$$

$$[H^{+}]$$ = 2 × 10-13M

pH = -log[H+]

pH = -log[2 × 10-13]

pH = 12.70

Answered by Abhisek | 1 year ago

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