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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 49 of 73
Marks: +1, -0
Calculate the pH of the following solutions :
(a) 2 g of TlOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)2\mathrm{Ca(OH)_2} dissolved in water to give 500 mL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
Solution:  
(a) Molar mass of TlOH = 221.4 g mol−1\text{g mol}^{-1}
∴ nTlOHn_{\mathrm{TlOH}} = 2 g221.4 g mol−1\frac{2\,\mathrm{g}}{221.4\,\mathrm{g\,mol}^{-1}} = 9.03 × 10−3 mol10^{-3}\,\mathrm{mol}
[OH−][\mathrm{OH}^-] = [TlOH] = 9.03×10−3 mol2 L\frac{9.03 \times 10^{-3}\,\mathrm{mol}}{2\,\mathrm{L}} = 4.51 × 10−3 M10^{-3}\,\mathrm{M}
pOH = – log(4.51 × 10−310^{-3}) = 2.35 and pH = 14 – 2.35 = 11.65
(b) nCa(OH)2n_{\mathrm{Ca(OH)_2}} = 0.3 g74 g mol−1\frac{0.3\,\mathrm{g}}{74\,\mathrm{g\,mol}^{-1}} = 4.05 × 10−3 mol10^{-3}\,\mathrm{mol}
[OH−][\mathrm{OH}^-] = 2[Ca(OH)2]2[\mathrm{Ca(OH)_2}] = 2 × 4.05×10−3 mol0.5 L\frac{4.05 \times 10^{-3}\,\mathrm{mol}}{0.5\,\mathrm{L}} = 1.62 × 10−2 M10^{-2}\,\mathrm{M}
pOH = - log (1.62 × 10−210^{-2}) = 1.79 and pH = 14 - 1.79 = 12.21
(c) nNaOHn_{\mathrm{NaOH}} = 0.3 g40 g mol−1\frac{0.3\,\mathrm{g}}{40\,\mathrm{g\,mol}^{-1}} = 7.5 × 10−3 mol10^{-3}\,\mathrm{mol}
[OH−][\mathrm{OH}^-] = [NaOH] = 7.5×10−3 mol0.2 L\frac{7.5 \times 10^{-3}\,\mathrm{mol}}{0.2\,\mathrm{L}} = 0.0375 M
pOH = – log(0.0375) = 1.43 and pH = 14 – 1.43 = 12.57
(d) M1V1M_1V_1 = M2V2M_2V_2
13.6 mol L−1\mathrm{mol\,L}^{-1} × 1 mL = M × 1000 mL
MHClM_{\mathrm{HCl}} = 0.0136 mol L−1\mathrm{mol\,L}^{-1}
[H+][\mathrm{H}^+] = [HCl] = 0.0136 M
∴ pH = – log(0.0136) = 1.87
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