Redox and Equivalent ConceptHard

Question

A 0.5 g sample of KH2PO4 is titrated with 0.1 M NaOH. The volume of base required to do this is 25.0 ml. The reaction is represented as H2PO4− + OH− → HPO42− + H2O. The percentage purity of KH2PO4 is

(K = 39, P = 31)

Options

A.68%
B.34%
C.85%
D.51%

Solution

$n_{eq}KH_{2}PO_{4} = n_{eq}OH^{-}$

Or $\frac{w}{136} \times 1 = \frac{25 \times 0.1}{1000} \Rightarrow w = 0.34\text{ g}$

$\therefore\%\text{ Purity = }\frac{0.34}{0.5} \times 100 = 68\%$

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