CBSE Class 10 Answered

H₂SO₄ will dissociate as:

H₂SO₄   ?  2H+    +   SO4^2-

0.01 moles of H₂SO₄ will give 2 X 0.01 = 0.02 moles of H⁺.

KOH will dissociate as:

KOH ? K⁺ + OH^-

0.01 moles of KOH will give 0.01 moles of OH^-.

Out of 0.02 moles of H⁺ from H₂SO₄, 0.01 moles of H⁺ will be neutralized by 0.01 moles of OH^- from KOH. The number of moles of OH^- remaining in 2 L water will be:

 0.02 - 0.01 = 0.01

Concentration of H⁺ = 0.01moles / 2L = 0.005 moles/L = 5 x 10^-3 moles/L

(2)

Ca(OH)₂  ?  Ca²⁺ + 2 OH^-
[OH^-] = 2 x 10^-8 M
[OH^-] by the auto ionization of water = 1 x 10^-7
total [OH^-] = 1 x 10^-7 + 2 x 10^-8 = 1.2 x 10^-7 M
pOH = - log 1.2 x 10^-7 = 6.9
pH = 14 - pOH = 7.1