50 mL of a solution containing 1 g each of Na2CO3, NaHCO3 and NaOH was titrated with N-HCl. What will be the titre reading when only phenolphthalein is used as indicator?

Question

50 mL of a solution containing 1 g each of Na2CO3, NaHCO3 and NaOH was titrated with N-HCl. What will be the titre reading when only phenolphthalein is used as indicator? (A) 35 mL (B) 32.5 mL (C) 24.5 mL (D) 34.4 mL

Tardigrade Admin · Accepted Answer

At the end point using phenolphthalein as indicator uses
complete

N a O H

and

\frac{1}{2}

Meq. of

N a_{2} C O_{3}

.

∴

Meq.of

N a O H + \frac{1}{2}

meq. of

N a_{2} C O_{3} =

Meq. of

H Cl

\frac{1}{40} \times 1000 + \frac{1}{2} + \frac{1}{106} \times 2 \times 1000 = V \times 1

∴ V = 34.4 m l

Q. $50 m L$ of a solution containing $1 g$ each of $N a_{2} C O_{3}, N a H C O_{3}$ and $N a O H$ was titrated with $N - H Cl$ . What will be the titre reading when only phenolphthalein is used as indicator?

$35 m L$

$32.5 m L$

$24.5 m L$

$34.4 m L$

Q. 50mL of a solution containing 1g each of Na2​CO3​,NaHCO3​ and NaOH was titrated with N−HCl. What will be the titre reading when only phenolphthalein is used as indicator?

35mL

32.5mL

24.5mL

34.4mL

At the end point using phenolphthalein as indicator uses complete NaOH and 21​ Meq. of Na2​CO3​. ∴ Meq.of NaOH+21​meq. of Na2​CO3​= Meq. of HCl 401​×1000+21​+1061​×2×1000=V×1 ∴V=34.4ml

Q. $50 m L$ of a solution containing $1 g$ each of $N a_{2} C O_{3}, N a H C O_{3}$ and $N a O H$ was titrated with $N - H Cl$ . What will be the titre reading when only phenolphthalein is used as indicator?

$35 m L$

$32.5 m L$

$24.5 m L$

$34.4 m L$