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The number of moles of `Ca(OH)_(2)` that must be dissolved to make 250 mL solution in water of pH = 10.65 isA. `5.6 xx 10^(-5)`B. `6.5 xx 10^(-5)`C. `4.5 xx 10^(-5)`D. `5.4 xx 10^(-5)` |
Answer» Correct Answer - A We know that pH + pOH = 3.35 pOH = (14-10.65) = 3.35 `[OH^(-)] = 10^(-3.35) = 4.47 xx 10^(-4) M` One molecule of `Ca(OH)_(2) = (4.47 xx 10^(-4))/(2) = 2.235 xx 10^(-4) M` No. of moles in `250 mL = (2.235 xx 10^(-4))/(4) = 5.6 xx 10^(-5)` |
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