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)\u00b725\u00b710^{-3}(L) = 1.25\u00b710^{-3}" mol.
Therefore, the number of the moles of KOH reacted is:
"n(KOH) = n(HCl) = 1.25\u00b710^{-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\u00b710^{-3} (mol)}{45\u00b710^{-3}(L)} = 0.028" M.
The mass of KOH in the solution is:
"m = nM = 1.25\u00b710^{-3} (mol)\u00b756.11(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}\u00b7250 = \\frac{0.070}{45}\u00b7250 = 0.39" g.
The percentage purity of KOH is:
"w = \\frac{0.39}{4} \u00b7100\\%=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 \u00b710^{-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.
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