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Question

What approximate volume of 0.40 M Ba(OH)2 must be added to 50.0 mL of 0.30 M NaOH to get a solution in which the molarity of the OH ions is 0.50 M?

A
33 mL
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B
66 mL
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C
133 mL
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D
100 mL
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Solution

The correct option is B 33 mL
Let us suppose that x mL of 0.40 M Ba(OH)2 soon is required to be added.

Assuming that Ba(OH)2 dissociates completely, number of millimoles of OH in solution =x×0.80 mmol

(Since each formula unit of Ba(OH)2 gives 2 OH ions.)

Number of millimoles of OH due to dissociation of NaOH =50×0.30 =15 mmol

Total number of millimoles of OH in solution=15+x×0.80 mmol

Total number of millimoles of OH required for 0.50 M solution =(50+x)mL×0.50 mmol/mL

=25+x×0.50 mmol

Hence, 15+0.80×x=25+x×0.50

0.30x=10

x=100.30=33.33

Approximately 33 mL of the Ba(OH)2 solution needs to be added.

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