Question

What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution

Answer 1

Answer: The $[OH^-]$ of a solution is $10^{-12}$ M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $HCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.0025\times 1000}{250}=0.01$

pH or pOH is the measure of acidity or alkalinity of a solution.

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.01$ moles of $HCl$ gives =$\frac{1}{1}\times 0.01=0.01$ moles of $H^+$

Putting in the values:

$[H^+][OH^-]=10^{-14}$

$[0.01][OH^-]=10^{-14}$

$[OH^-]=10^{-12}$

Thus the $[OH^-]$ of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is $10^{-12}$ M