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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 48 of 73
Marks: +1, -0
Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
Solution:  
(a) HCl + aq → H++Cl−\mathrm{H}^{+} + \mathrm{Cl}^{-}
∴ [H+][\mathrm{H}^{+}] = [HCl] = 3 × 10−310^{-3} M, pH = - log (3×10−3)(3 \times 10^{-3}) = 2.52
(b) NaOH + aq → Na++OH−\mathrm{Na}^{+} + \mathrm{OH}^{-}
∴ [OH−][\mathrm{OH}^{-}] = 5 × 10−3 M10^{-3}\,\text{M}
∴ [H+][\mathrm{H}^{+}] = 10−45×10−3\frac{10^{-4}}{5 \times 10^{-3}} = 2 × 10−12 M10^{-12}\,\text{M} , pH = - log (2 × 10−1210^{-12}) = 11.70
(c) HBr + aq → H++Br−\mathrm{H}^{+} + \mathrm{Br}^{-}
∴ [H+][\mathrm{H}^{+}] = 2 × 10−310^{-3} M, pH = - log (2×10−3)(2 \times 10^{-3}) = 2.70
(d) KOH + aq → K++OH−\mathrm{K}^{+} + \mathrm{OH}^{-}
∴ [OH−][\mathrm{OH}^{-}] = 2 × 10−3 M10^{-3}\,\text{M} , [H+][\mathrm{H}^{+}] = 10−142×10−3\frac{10^{-14}}{2 \times 10^{-3}} = 5 × 10−1210^{-12}
pH = - log (5×10−12)(5 \times 10^{-12}) = 11.30
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