Question

A solution was made by dissolving 1.77 g Ba(OH)2 in 100. mL final volume. (Note: For the purpose of significant figures, assume Kw = 1.00 x 10-14) (a) What is the molar concentration of OH- in the solution?

Answer 1

Answer: The molar concentration of $OH^-$ in the solution is 0.2 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 =$\frac{\text {given mass}}{\text {molar mass}}=\frac{1.77g}{171g/mol}=0.010$

$V_s$ = volume of solution in ml = 100 ml

$Molarity=\frac{0.010\times 1000}{100}=0.1M$

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

As 1 mole of $Ba(OH)_2$ gives 2 moles of $OH^-$

Thus 0.1 moles of $Ba(OH)_2$ gives = $\frac{2}{1}\times 0.1=0.2$ moles of $OH^-$

The molar concentration of $OH^-$ in the solution is 0.2 M