Question

What volume of $0.2MKOH$ will be required to neutralize $100$ mL of $0.1M{H}_{3}P{O}_4$ using methyl red indicator (change of colour pink $\to$ yellow) followed by the addition of bromothymol blue indicator?

• A. $50$ mL
• B. $100$ mL
• C. $150$ mL
• D. $200$ mL

Answer 1

The correct option is B $100$ mL

Methyl red indicates the first step ionization of $H_{2}P{O}_4$.

Bromothymol blue indicates the second step ionization of $H_{3}P{O}_4$.
$H_{2}P{O}_4^{⊝}\to H^{⨁}+HP{O}_4^{2-}(n=1)$

First case: When methyl red is added.
$H_{3}P{O}_4\to H_{2}P{O}_4^{⊝}+H^{⨁}(n=1)(N=M\times 1)$
$KOH\equiv H_{3}P{O}_4$
$N_1{V}_1\equiv N_2{V}_2$
$0.2\times 1\times V_1=0.1\times 1\times 100$
$V_1=50$ mL

Second case: When bromothymol blue is added.
$H_{2}P{O}_4^{⊝}\to H^{⨁}+HP{O}_4^{2-}(n=1)(N=M\times 1)$
$KOH\equiv H_{2}P{O}_4^{⊝}$
$N_1{V}_1\equiv N_2{V}_2$
$0.2\times 1\times V_1=0.1\times 1\times 100$
$V_1=50$ mL
Total volume $=50+50=100$ mL