NCERT Class XI Chemistry Equilibrium Solutions

Question : 57

Total: 73

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?

Solution:

Molar mass of KOH = 56.0 gmol–1
∴ No. of moles of KOH =

0.561g

56.0gmol−1

= 0.01 mol
and conc. of KOH =

0.01

0.2L

mol = 0.05 mol/L
KOH(aq) → K(aq)++OH(aq)−
[K+] = [OH−] = 0.05 M and [H+] =

[OH−]

1.0×10−14

0.05

= 2.0 × 10−13
∴ pH = - log (2.0 × 10−13) = 12.70

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