Question

$2.5g$ of the carbonate of a metal was treated with $100ml$ of $1N{H}_{2}S{O}_4$. After the completion of the reaction, the solution was boiled off to expel $C{O}_2$ and was then titrated against $1NNaOH$ solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is $20$

• A. 50 mL
• B. 25 mL
• C. 75 mL
• D. 100 mL

Answer 1

Answer: (A)

Equivalent weight of metal carbonate

$=20+30=50$

$2.5g$ of metal carbonate $=\frac{2.5}{50}=0.05eq$

Number of equivalent of $H_{2}S{O}_4$ would have reacted $=0.05$

Number of equivalent of $H_{2}S{O}_4$ taken

$=\frac{100\times 1}{1000}=0.1$

Number of equivalent of $H_{2}S{O}_4$ remains unreacted

$=0.1-0.05=0.05eq$

$\therefore$ Number of equivalent of alkali consumed $=0.05eq$

milli eq. $=$ Normality $\times$ Volume in $mL$

$\therefore 1.0\times V=0.05\times 1000$

$V=\frac{0.05\times 1000}{1.0}=50ml$