Question

What approximate volume of O 40 M Ba(OH), must be added to 50.0 mL of 0.30 M NaOH to get a solution in
which the molarity of the Oitions is 0.50 M?
(A) 33 mL
(B) 66 mL
(0) 133 mL
(D) 100 ml​

Answer 1

Answer:

a) 33.3% mL

Explanation:

Let the quantity of 0.40 M Ba(OH)2 to be added = x

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

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 =10/0.30

= 33.33

Therefore, 33.3% mL of the Ba(OH)2 solution needs to be added to the solution.