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Stoichiometric Calculations

Titration of ...

Question

Titration of a $0.7439$ g sample of impure $N a_{2} B_{4} O_{7} .10 H_{2} O$ (borax) requires $31.64$ mL of $0.108 M H C l$ for the reaction. The result of this analysis in terms of percentage is (in nearest integer) :
(Molecular weight of $B = 11$ g/mol)

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Solution

Number of moles of $H C l$ $= \frac{0.108 \times 31.64}{1000} = 0.00342$ moles

$1$ mole borax corresponds to $2$ moles of $H C l$ .

Number of moles of borax $= \frac{0.00342}{2} = 0.00171$ moles

Molar mass of $N a_{2} B_{4} O_{7} .10 H_{2} O = 202 + 180 = 382$ g/mol

Percentage of borax $= \frac{0.00171 \times 382}{0.7439} \times 100 = 87.73 %$ $\sim 88 %$

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