Question

$56.0$ g $KOH$, $138.0$ g $K_{2}C{O}_3$ and $100.0$ g $KHC{O}_3$ is dissolved in water and the solution is made $1L$. $10$ mL of this stock solution is titrated with $2.0M$ $KCl$. Which of the following statements is/are correct?

• A. When phenolphthalein is used as an indicator from the very beginning, the titre value of $HCl$ will be $60$ mL.
• B. When phenolphthalein is used as an indicator from the very beginning, the titre value of $HCl$ will be $40$ mL.
• C. When methyl orange is used as an indicator from the very beginning, the titre value of $HCl$ will be $80$ mL.
• D. When methyl orange is used as an indicator after the first end point, the titre value of $HCl$ will be $40$ mL.

Answer 1

Answer: (B), (C), (D)

The correct options are
B When phenolphthalein is used as an indicator from the very beginning, the titre value of $HCl$ will be $40$ mL.
C When methyl orange is used as an indicator after the first end point, the titre value of $HCl$ will be $40$ mL.
D When methyl orange is used as an indicator from the very beginning, the titre value of $HCl$ will be $80$ mL.

$a.$

mEq of $KOH/L=\frac{56}{56/1}\times {10}^3=1000=20/20mL$ solution.
mEq of $K_{2}C{O}_3/L=\frac{138}{138/2}\times {10}^3=2000=40/20mL$ solution.
mEq of $KHC{O}_3/L=\frac{100}{100/1}\times {10}^3=1000=20/20mL$ solution.
$b.$ With phenolphthalein:

mEq of $KOH+\frac{1}{2}$ mEq of $K_{2}C{O}_3$=mEq of $HCl$
$20+\frac{40}{2}=V\times 1N(n=1)$
$V_{HCl}=40$ mL
$c.$ With methyl orange:

mEq of $KOH+$ mEq of $K_{2}C{O}_3+$ mEq of $KHC{O}_3$= mEq of $HCl$
$20+40+20=V\times 1N$
$V_{HCl}=80$ mL
$d.$ With methyl orange after the first end point:

$\frac{1}{2}$ mEq of $K_{2}C{O}_3+$ mEq of $KHC{O}_3$=mEq of $HCl$
$20+20=V\times 1N$
$V_{HCl}=40$ mL

Hence options $B,C$ & $D$ are correct.