Question

1. 45cm3 of a solution containing 4g of impure potassium hydroxide in 250cm3 was neutralized by 25cm3 of 0.05M hydrochloric acid solution. Calculate
i. The concentration of the pure KOH in moles/dm3
ii. The percentage purity of KOH
iii. The percentage impurity of KOH
iv. The number of moles of hydroxyl ions present in the pure KOH that reacted

Answer 1

The balanced reaction equation is:

KOH + HCl $\rightarrow$ H₂O + KCl.

As one can see, 1 mol of potassium hydroxide reacts with 1 mol of hydrochloric acid:

$n(KOH) = n(HCl)$ .

The number of the moles of HCl reacted is the product of the concentration and the volume of the HCl solution:

$n(HCl) = cV = 0.05(M)·25·10^{-3}(L) = 1.25·10^{-3}$ mol.

Therefore, the number of the moles of KOH reacted is:

$n(KOH) = n(HCl) = 1.25·10^{-3}$ mol.

The concentration of the 45 cm3 solution that contains 1.25·10^-3 mol of KOH is:

$c(KOH) = \frac{n}{V} = \frac{1.25·10^{-3} (mol)}{45·10^{-3}(L)} = 0.028$ M.

The mass of KOH in the solution is:

$m = nM = 1.25·10^{-3} (mol)·56.11(g/mol) = 0.070$ g.

The mass of KOH in 45 mL is proportional to the mass of KOH in 250 mL:

$\frac{m_1}{45} =\frac{m_2}{250}$

$m_2 =\frac{m_1}{45}·250 = \frac{0.070}{45}·250 = 0.39$ g.

The percentage purity of KOH is:

$w = \frac{0.39}{4} ·100\%=9.7$ %.

The percentage impurity of KOH is:

$w_i = 100-9.7 = 90.3$ %.

As KOH is a strong electrolyte, the number of the moles of hydroxide anions is equal to the number of the moles of KOH:

$n(KOH) = n(OH^-) = 1.25 ·10^{-3}$ mol.

Answer:

i. The concentration of the pure KOH is 0.028 moles/dm³

ii. The percentage purity of KOH is 9.7%

iii. The percentage impurity of KOH is 90.3 %

iv. The number of moles of hydroxyl ions present in the pure KOH that reacted is 1.25·10^-3 mol.

P.S. I suppose that there is a mistake in the concentration of HCl used. If it was 0.5 M and not 0.05 M, the purity of KOH would be 97%, the percentage impurity would be 3%, the concentration of the pure KOH would be 0.28 moles/dm³ and the number of moles of hydroxyl ions present in the pure KOH that reacted would be 1.25·10^-2 mol.