Answer to Question #122738 in Inorganic Chemistry for Frank

Question #122738
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
1
Expert's answer
2020-06-17T08:03:28-0400

The balanced reaction equation is:

KOH + HCl H2O + KCl.

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

.

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

mol.

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

mol.

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

M.

The mass of KOH in the solution is:

g.

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

g.

The percentage purity of KOH is:

%.

The percentage impurity of KOH is:

%.

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

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.


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