Question

# A mixture of NaOH and Na2CO3 solution is titrated with 50 mL of N10 HCl using phenolpthalein. At this stage methyl orange was added and addition of acid was continued. The second end point was reached after further addition of 10 mL of N10 HCl . The amount of NaOH (in g) in the solution is:

A
3.2 g
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B
0.16 g
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C
0.08 g
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D
0.4 g
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Solution

## The correct option is B 0.16 g1st titration Meq. of NaOH+12 Meq. of Na2CO3=Meq. of HCl=50×110=5 2nd titration 12 Meq. of Na2CO3=Meq. of HCl=10×110=1 ∴ Meq. of NaOH remains=5−1=4 Wteight of NaOH=Meq. of NaOH×Equivalent weight1000 =4×401000 g =0.16 g

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