Question

a g $KH{C}_2{O}_4$ is used to neutralize 100 mL of 0.02M $KMn{O}_4$ in acid medium, where as b g $KH{C}_2{O}_4$ is used to neutralize 100 mL of 0.02M $Ca(OH{)}_2$ then, then,

• A. $a=b$
• B. $4a=5b$
• C. $5a=4b$
• D. $2a=b$

Answer 1

Answer: (B)

milli eq. of $KH{C}_2{O}_4=$ milli eq. of $KMn{O}_4$ ${\left(C^{3+}\right)}_2\to 2C^{4+}+2e$

$M{n}^{7+}+5e\to M{n}^{2+}$

$\frac{a}{\frac{M}{2}}\times 1000=100\times 0.02\times 5$

$a=5\times 10^{-3}M$

milli eq. of $KH{C}_2{O}_4=$ milli eq. of $Ca(OH{)}_2$

$KH{C}_2{O}_4$ acts as acidic salt $\frac{b}{M}\times 1000=100\times 0.02\times 2$

$b=4\times 10^{-3}M$

$\frac{a}{b}=\frac{5}{4}$