Question

a g $KHC_{2}O_{4}$ is used to neutralize 100 mL of 0.02M $KMnO_{4}$ in acid medium, where as b g $KHC_{2}O_{4}$ 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 $KHC _{2} O _{4}=$ milli eq. of $KMnO _{4}$ $\left( C ^{3+}\right)_{2} \rightarrow 2 C ^{4+}+2 e$

$Mn ^{7+}+5 e \rightarrow Mn ^{2+}$

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

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

milli eq. of $KHC _{2} O _{4}=$ milli eq. of $Ca ( OH )_{2}$

$KHC _{2} O _{4}$ acts as acidic salt $\frac{b}{M} \times 1000=100 \times 0.02 \times 2$

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

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