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Question

A solution containing Na2CO3 and NaOH requires 300ml of 0.1NHCl using phenolphthalein as indicator. Methyl orange is then added to the above solution when 25ml of 0.2NHCl is required. The amount of NaOH present in the solution is:

A
1.0 g
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B
2.0 g
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C
1.5 g
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D
0.2 g
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Solution

The correct option is C 1.0 g
The reaction are as follows:
NaOH+HClNaCl+H2O
Na2CO3+HCl2NaCl+CO2+H2O
Phenolphthalein changes its color when the solution becomes neutral i.e. when all the base is consumed and half of the sodium carbonate is converted to sodium hydrogen carbonate.
So, we can say, 300 ml HCl of 0.1N HCl neutralizes the entire amount of NaOH and half of Na2CO3.
As given, 50ml of 0.1N (which is equal to 25 ml of 0.2N) of HCl is utilized to neutralize the remaining half of Na2CO3.
So, 250 ml (30050 ml) of 0.1N of HCl is required to neutralize NaOH completely.
As number of milliequivalents of HCl= number of milliequivalents of NaOH
250×0.1 meq of HCl=250×0.1 meq of NaOH which is equal to 25 milimoles of NaOH, which in turn, is equal to 1 g of NaOH.

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