Question

What volume of a 0.1 M NaOH solution must be combined with 25 mL of 0.2 M HNO3 to reach a pH of 11.5?

Answer 1

Let V ml of NaOH be added

we know that

moles = molarity x volume (ml) / 1000

so

moles of HN03 = 0.2 x 25 x 10-3 = 5 x 10-3

moles of NaOH = 0.1 x V x 10-3 = 0.1V x 10-3

now

the reaction is

NaOH + HN03 ---> NaN03 + H20

we know that

NaOH --> basic

HN03---> acidic

NaN03 --> neutral

given the final pH is 11.5

so

the final solution is basic

so

only

NaOH should remain in the final solution

So

NaOH is in excess

moles of NaOH reacted = moles of HN03 added = 5 x 10-3

unreacted moles of NaOH = ((0.1V x 10-3 ) - 5 x 10-3

final volume = 25 + V

so

conc of NaOH = ( 0.1V x 10-3 - 5 x 10-3 ) x 1000 / ( 25 + V)

[NaOH] = (0.1V - 5) / ( 25 + V)

now

we know that

pH = -log [H+]

so

11.5 = -log [H+]

[H+] = 3.16228 x 10-12

we know that

[H+] [OH-] = 10-14

so

3.16228 x 10-12 x [OH-] = 10-14

[OH-] = 3.16228 x 10-3

NaOH ---> Na+ + OH-

so

in the final solution

[NaOH] = [OH-] = 3.16228 x 10-3

so

(0.1V - 5) / ( 25 + V) = 3.16228 x 10-3

(0.1V - 5) = 3.16228 x 10-3 ( 25 + V)

V = 52.45

so

52.45 ml of NaOH solution is combined