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Standard XII

Chemistry

Solubility Product

If 0.561 g of...

Question

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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Solution

$[K O H_{a q}] = \frac{0.561}{\frac{1}{5}} g / L = 2.805 g / l = 2.805 \times \frac{1}{56.11} M = 0.5 M K O H_{(a q)} \to K_{(a q)}^{+} + O H_{(a q)}^{-} [O H^{-}] = 0.5 M = [K^{+}] [H^{+}] [H^{-}] = K_{w} [H^{+}] \frac{K_{w}}{[O H^{-}]} = \frac{10^{- 14}}{0.05} = 2 \times 10^{- 13} M ∴ p H = 12.70$

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Solubility Product

Standard XII Chemistry

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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. What is its pH?

[KOHaq]=0.56115g/L=2.805g/l=2.805×156.11M=0.5MKOH(aq)→K+(aq)+OH−(aq)[OH−]=0.5M=[K+][H+][H−]=Kw[H+]Kw[OH−]=10−140.05=2×10−13M∴pH=12.70