20 g of $B a (O H)_{2}$ is dissolved in 500 mL of water. 25 mL of this is used to titrate a $H_{2} S O_{4}$ solution. On titration it gave a titre value of 35 mL. Calculate the molarity of $H_{2} S O_{4}$ solution.
(Molar mass of Ba = 137 g/mol)

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Solution

Molarity of $B a (O H)_{2}$ solution = $\frac{20}{171} \times 2 = \frac{40}{171} M$

So, the reaction involved in titration is:
$B a (O H)_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H_{2} O$

Thus 1 mol of the base ( $B a (O H)_{2}$ ) reacts with with one mole of the acid ( $H_{2} S O_{4}$ )

Hence we can write, $M_{1} V_{1} = M_{2} V_{2}$

Where, $M_{1}$ = molarity of $B a (O H)_{2}$
$V_{1}$ = volume of $B a (O H)_{2}$
$M_{2}$ = molarity of $H_{2} S O_{4}$
$V_{2}$ = volume of $H_{2} S O_{4}$

$∴ \frac{40}{171} \times 25 = M_{2} \times 35$
$\Rightarrow M_{2} = \frac{40 \times 25}{171 \times 35} = 0.167 M$

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