` 1 g` of impure `Na_(2)CO_(3)` is dissolved in water and the solution is made upto `250 mL`. To `50 mL` of this solution, `50 mL` of `0.1 N HCl` is added and the mixture after shaking well required `10 mL` of `0.16 N NaOH` solution for complete neutralization. Calculation percent purity of the sample of `Na_(2) CO_(3)`.

` 1 g` of impure `Na_(2)CO_(3)` is dissolved in water and the solution

1.	` 1 g` of impure `Na_(2)CO_(3)` is dissolved in water and the solution is made upto `250 mL`. To `50 mL` of this solution, `50 mL` of `0.1 N HCl` is added and the mixture after shaking well required `10 mL` of `0.16 N NaOH` solution for complete neutralization. Calculation percent purity of the sample of `Na_(2) CO_(3)`.
Answer» `1.0 g` of (impure) `Na_(2) CO_(3) + H_(2)O rarr 250 mL` `(50 mL of Na_(2)CO_(3) + 50 mL of 0.1 N NaOH)` `= 10 mL of 0.16 N NaOH` In this question, `HCl` is in excess. `implies` Excess mEq of `HCl = mEq of NaOH = 0.16 xx 10 = 1.6` mEq of `HCl` added to `Na_(2) CO_(3) = 0.1 xx 50 = 5` `implies` mEq of `HCl` used to neutralised `Na_(2) CO_(3) = 5 - 1.6 = 3.4` So, mEq of `Na_(2)CO_(3)` (pure) in `50 mL = 3.4` `implies` mEq of pure `Na_(2) CO_(3)` is `250 mL` `3.4 xx (250)/(50) = 17` or `("Weight")/(Ew) xx 100 = 17` `implies "Weight = (17 xx (106 // 2))/(1000) = 0.901 g` `(Na_(2) CO_(3)` is diacidic base `implies Ew = 106//2)` So mass of pure `Na_(2)CO_(3) = (0.901)/(1) xx 100 = 90.1%`

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