Question

What will be the normality of the solution obtained by mixing $50mL$ of $0.2M{H}_{2}S{O}_4$​ with $50mL$ of $0.1NCa(OH{)}_2$ ?

• A. $0.20N$
• B. $0.20N$
• C. $0.15N$
• D. $0.10N$

Answer 1

The correct option is C $0.15N$

$\text{Normality}=\text{Molarity}\times n_f$
$H_{2}S{O}_4\text{is dibasic acid}\therefore n-factor=2$
$N_{1}of{H}_{2}S{O}_4=0.2\times 2=0.4N$
$N_{2}ofCa(OH{)}_2=0.1N$

The required formula is -
$N_1{V}_1-N_2{V}_2=N_3{V}_3$​

where, $V_1$​ and $V_2$​ are the volumes of $H_{2}S{O}_{4}andCa(OH{)}_2$ respectively.
Given , $V_1=V_2=50mL$
$V_3=50+50=100mL$
Where, $N_3$​ and $V_3$​ are the normality & volume of the final solution.

$(0.4\times 50)-(0.1\times 50)=N_3\times \left(100\right)N_3=\frac{15}{100}{N}_3=0.15N$
The normality of the resultant acidic solution is $0.15N$.

Important list of Reactions and Half Reactions :

$C{r}_2{O}_7^{2-}⟶C{r}^{3+}$

$Mn{O}_4^{-}⟶M{n}^{2+}\left(Acidic\right)$

$Mn{O}_4^{-}⟶Mn{O}_2\left(Basic/Neutral\right)$

$Mn{O}_4^{-}⟶Mn{O}_4^2-\left(HighlyBasic\right)$

$I{O}_3^{-}⟶I^{-}$

$Mn{O}_2⟶M{n}^{2+}\left(Acidic\right)$

$I_2⟶I^{-}$

$Cl{O}^{-}⟶C{l}^{-}$

$H_2{O}_2⟶H_{2}O$

$N{O}_3^{-}⟶N{O}_2^{-}$

$N{O}_3^{-}⟶N{O}_2$

$N{O}_3^{-}⟶NO$

$N{O}_3^{-}⟶N_{2}O$

$N{O}_3^{-}⟶N{H}_4^{+}$

\(O_3 \longrightarrow O_2 + O^{2-}

$F{e}^{2+}⟶F{e}^{3+}$

$S{n}^{3+}⟶S{n}^{3+}$

$S_2{O}_3^{2-}⟶S_4{O}_6^{2-}$

$S^{2-}⟶S$

$N{O}_2^{-}⟶N{O}_3^{-}$

$H_2{O}_2⟶O_2$

$C_2{O}_4^{2-}⟶C{O}_2$

$Zn⟶Zn{O}_2\left(Basic\right)$