Question

What volume of 0.40 M Ba(OH) 2 must be added to 50 mL of 0.30 M NaOH to give a solution
0.50 M in OH − ?

Answer 1

Answer:

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Answer 2

Answer:

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

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

⇒0.30⇒x=10

x=

0.30

10

=33.33

Approximately 33 mL of the Ba(OH)

2

solution needs to be added