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Question

20 g a sample of Ba(OH)2 is dissolved in 10 mL of 0.5 N HCL solution: The excess of HCL was titrated with 0.2 N NaOH. The volume of NaOH used was 20 cc. Calculate the percentage of Ba(OH)2 in the sample.

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Solution

Excess HCl left after reaction with Ba(OH)2=

20 mL 0.2 N NaOH=x mL 0.5 N HCl

V1N1=V2N2

20×0.2=x×0.5

x=8 mL

Only 108=2 mL HCl reacted with the Ba(OH)2 sample.

2 mL 0.5 N HCl=2×0.51000=0.001 moles

Now, Ba(OH)2+2HClBaCl2+H2O

Ratio is 1 : 2

x : 0.001

x=0.0005 moles

Weight of Ba(OH)2 in sample

=molar weight×0.0005

=171×0.0005 g=0.0855 g

%Ba(OH)2 in sample = weight of Ba(OH)2 in the sampleweight of the sample×100

=0.085520×100

=0.4275 %


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