$2.5 g$ of a mixture containing ${CaCO}_{3}, Ca {({HCO}_{3})}_{2}$ and $NaCl$ was dissolved in $100 m L$ water and its $10 m L$ portion required $10 m L 0.05 M$ $H_{2} {SO}_{4}$ solution to reach the phenolphthalein end Point Another $10 m L$ portion of the same stock solution required $32.35 m L$ of the same acid solution reach the methyl orange end point. Determine mass percentage ratio of ${CaCO}_{3}$ and $Ca ({HCO}_{3})_{2}$ in the original mixture.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here,

For phenolphthalein end point,

$\begin{array}{l} \underset{1}{2 {CaCO}_{3}} + \underset{\underset{= 0.5}{10 \times 0.05}}{H_{2} {SO}_{4}} ⟶ Ca {({HCO}_{3})}_{2} + {CaSO}_{4} \end{array}$

For methyl orange end point

$\underset{1}{{CaCO}_{3}} + \underset{1 m m o l}{H_{2} {SO}_{4}} ⟶ {CaSO}_{4} + H_{2} O + {CO}_{2}$

$Ca {({HCO}_{3})}_{2} \underset{\begin{matrix} (32.35 \times 0.05 - 1) \\ = 6.175 \times 10^{- 1} \end{matrix}}{+ H_{2} {SO}_{4}} ⟶ {CaSO}_{4} + 2 H_{2} O + 2 {CO}_{2}$ $Ratio = \frac{1 \times 100}{6.175 \times 10^{- 1} \times 162} = 1$