Question

74g ca(oh)2 is dissolved in water to give 200ml solution at 298 k. The hydrogen ion concentration in the solution is

Answer 1

Answer: $10^{-13}M$

Explanation:

$\text{no of moles}=\frac{\text {Given mass}}{\text {Molar mass}}$

$\text{no of moles of calcium hydroxide} =\frac{0.74g}{74g/mol}=0.01mol$

$Molarity=\frac{\text {Moles of calcium hydroxide}}{\text {Volume of solution in L}}$

$Molarity=\frac{0.01mole}{0.2L}=0.05M$

1 M of $Ca(OH)_2$ produces 2M of $OH^-$ ions.

0.05 M of $Ca(OH)_2$ produces=$\frac{2}{1}\times {0.05}=0.1M$of $OH^-$ ions.

$pOH=-log[OH^-]$

$pOH=-log[0.1M]=1$

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1 = 13.

$pH=-log[H^+]$

$13=-log[H^+]$

$[H^+]=10^{-13}M$