Name: Understanding Redox Titrations
Uploaded: 2019-07-19
Duration: 11 min 41 s
Description: Redox titration is a great way to identify whether a reaction is completed or not, where we know that reaction is complete, and it helps us find the strength...

Question

One

g

of impure sodium carbonate is dissolved in water and the solution is made up to

250 mL

. To

50 mL

of this made up solution,

50 mL

of

0.1 N -

HCl

is added and the mix after shaking well required

10 mL

of

0.16 N - NaOH

solution for complete titration. Calculate the

%

purity of the sample.

Accepted Answer

Given: Weight of impure sodium carbonate $= 1 gm, 50 mL of 0.1 N HCl and 10 mL of 0.16 N NaOH$

${Na}_{2} {CO}_{3} + HCl \to NaCl + H_{2} O + {CO}_{2}$

$HCl + NaOH \to NaCl + H_{2} O$

By the law of equivalence

Meq. of ${Na}_{2} {CO}_{3} = Meq$ of $HCl$ used

$Meq . of HCl used = Meq . of HCl with {Na}_{2} {CO}_{3} - Meq . of HCl with NaOH$

$Meq . = \frac{weight}{equivalent weight} \times 1000 = normality \times volume$

$\Rightarrow \frac{w}{\frac{106}{2}} \times 1000 = (50 \times 0.1 - 10 \times 0.16) \times \frac{250}{50}$

$\Rightarrow w = 0.901 gm$

So $%$ purity of the sample $= \frac{0.901}{1} \times 100 = 90.1 %$

Hence, $%$ purity of the sample is $90.1 %$

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !

One g of impure sodium carbonate is dissolved in water and the solution is made up to 250 mL. To 50 mL of this made up solution, 50 mL of 0.1 N-HCl is added and the mix after shaking well required 10 mL of 0.16 N-NaOH solution for complete titration. Calculate the % purity of the sample.

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !