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Find the pH of the following solution:(a) 3.2 g of hydrogen chloride dissolved in 1.00 L of water.(b) 0.28 g of potassium hydroxide dissolved in 1.00 water. |
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Answer» (a) Molarity of HCl = \(\frac{3.2}{36.5}\) x 1 0.089 M = 8.9 x 10-2 Since HCl is completely ionized, [H+] = 8.9 x 10-2 mol L-1 pH = - [log H+] = -[log 8.9 x 10-2] = -[-2 + 0.7762] = 1.2238 (b) Molarity of KOH = \(\frac{0.28}{56\times1}\) = \(\frac{1}{200}\) = \(\frac{1}{2\times10^2}\) = 0.5 x 10-2 = 5 x 10-2 mol l-1 [H+] = \(\frac{K_w}{[OH^-]}\) [H+] = \(\frac{1.0\times10^{-14}}{5\times10^{-2}}\) = 0.2 x 10-11 = 2 × 10−12 pH = − log (2 × 10−12) = −(−12 + 0 ∙ 3010) = 11.699 |
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