100 ml of a mixture of NaOH and Na2SO4 is neutralised by 10 ml of 0.5 M H2SO4. Hence NaOH in 100 mL solution is

Question

100 ml of a mixture of NaOH and Na2SO4 is neutralised by 10 ml of 0.5 M H2SO4. Hence NaOH in 100 mL solution is (A) 0.2 g (B) 0.4 g (C) 0.6 g (D) none of these

Tardigrade Admin · Accepted Answer

As Na2SO4 is already neutral so only NaOH will be neutralized by H2SO4 ∴ Eq. of NaOH = Eq. ofH2SO4 (w×1000/40×100) ×1× (100/1000)=10×10-3×0.5 ×2 W=40 ×10-2 =0.4 g

Q. $100 m l$ of a mixture of $N a O H$ and $N a_{2} S O_{4}$ is neutralised by 10 ml of 0.5 M $H_{2} S O_{4}$ . Hence NaOH in $100$ mL solution is

$0.2$ g

$0.4$ g

$0.6$ g

none of these

As $N a_{2} S O_{4}$ is already neutral so only $N a O H$ will be neutralized by $H_{2} S O_{4}$
$∴ Eq . o f N a O H = Eq . o f H_{2} S O_{4}$
$\frac{w \times 1000}{40 \times 100} \times 1 \times \frac{100}{1000} = 10 \times 1 0^{- 3} \times 0.5 \times 2$
$W = 40 \times 1 0^{- 2} = 0.4 g$

Q. 100ml of a mixture of NaOH and Na2​SO4​ is neutralised by 10 ml of 0.5 M H2​SO4​. Hence NaOH in 100 mL solution is

0.2 g

0.4 g

0.6 g