Ncert Exercises & Solutions

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?

[{KOH}_{aq}] = \frac{0.561}{\frac{1}{5}} g / L

= 2.805 g / L

= 2.805 \times \frac{1}{56.11} M

= . 05 M

{KOH}_{(aq)} \to K_{(aq)}^{+} + {OH}^{-}_{(aq)}

[{OH}^{-}] = . 05 M = [K^{+}]

[H^{+}] [H^{-}] = K_{w}

[H^{+}] = \frac{K_{w}}{[{OH}^{-}]}

= \frac{10^{- 14}}{0.05} = 2 \times 10^{- 13} M

∴ pH = 12.70

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