Question

An aqueous solution is prepared by mixing 100.00 mL of 0.100 M NaNO2 and 35.00 mL of 0.200 M HNO2 at 25 °C. To the above solution is added 4.00 mL of 0.250 M KOH. Calculate the pH of the resultant solution at 25 °C. For HNO2, Ka = 4.47 x 10-4 at 25 °C.

Answer 1

we know that

moles = molarity x volume (L)

so

moles of NaN02 = 0.1 x 100 x 10-3 = 10 x 10-3

moles of HN02 = 0.2 x 35 x 10-3 = 7 x 10-3

now

moles of KOH added = 0.25 x 4 x 10-3 = 1 x 10-3

now

the reaction is

HN02 + OH- ---> H20 + N02-

so

moles of HN02 reacted = moles of KOH added = 1 x 10-3

so

moles of HN02 left = 7 x 10-3 - 1 x 10-3 = 6 x 10-3

now

moles of N02- formed = moles of KOH added = 1 x 10-3

so

new moles of N02- = 10 x 10-3 + 1 x 10-3 = 11 x 10-3

now

HN02 and N02- form a buffer solution

we know that

for buffers

pH = pKa + log [ conjugate base / acid ]

also

pKa = -log Ka

so

pH = -log Ka + log [N02- / HN02]

so

pH = -log 4.47 x 10-4 + log [ 11 x 10-3 / 6 x 10-3]

pH = 3.613

so

the pH of the resulting solution is 3.613