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`        5 mL of 8 N nitric acid, 4.8 mL of 5 N hydrochloric acid and certain volume of 17 M sulphuric acid are mixed together and made up to 2 L. 30 mL of this acid mixture exactly meutralise 42.9 mL of sodium carbonate solution containing one gram of Na2CO3 . 10H2O in 100 mL of water. Calculate the amount in gram of the sulphate ions in solution.`
3 years ago

Deepak Patra
474 Points
```										Sol. Molecular weight of Na2CO3 ∙ 10H2O = 286
⇒ Molarity of carbonate solution = 1/286   ×  1000/100 = 0.035
⇒ Normality of carbonate solution = 2 × 0.035 = 0.07 N
In acid solution : Normality of HNO3 = (8 ×5)/2000 = 0.02
Normality of HCI = (5 ×4.8)/2000 = 0.012
Let normality of H2SO4 in final solution be N.
⇒ (N + 0.02 + 0.012) × 30 = 0.07 × 42.9
⇒ N = 0.0681
⇒ Gram equivalent of 〖SO〗_4^(2-) in 2L solution = 2 × 0.0681 = 0.1362
⇒ Mass of 〖SO〗_4^(2-) in solution = 0.1362 = 96/2 = 6.5376 g

```
3 years ago
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