Question

A 25.0 mL sample of 0.175 M HClO(aq) is titrated with 0.100 M NaOH(aq). What is the pH of a solution after the addition of 25.0 mL of NaOH? (K_aof HClO = 3.5

Answer 1

the reaction is given by

HCl0 + NaOH ---> NaCl0 + H20

now

we know that

moles = molarity x volume (L)

so

moles of HCl0 taken = 0.175 x 25 x 10-3 = 4.375x 10-3

moles of NaOH added = 0.1 x 25 x 10-3 = 2.5 x 10-3

final volume = 25 + 25 = 50 ml

now

consider the reaction

HClO + NaOH ---> NaCl0 + H20

we can see that

moles of HClO reacted = moles of NaOH added = 2.5 x 10-3

moles of HClO remaining = initial - final = 4.375 x 10-3 - 2.5 x 10-3 = 1.875 x 10-3

moles of NaClO formed = mole sof HClO reacted = 2.5 x 10-3

now

only

HClO and NaClo are present in the solution

this is a combination of weak acid and salt of a weak acid

it forms a acidic buffer

for acid buffers

according to hasselbach hendersen equation

pH = pKa + log [ salt / acid]

also

pKa = -log Ka

so

pH = -log Ka + log [ NaClo/HCl0]

now

molarity = moles / volume (L)

as the final volume is same for both , they cancel out

so

ratio of molarities = ratio of moles

so

pH = -log 3.5 x 10-8 + log [ 2.5 x 10-3 / 1.875 x 10-3]

pH = 7.581

so the pH of the final solution is 7.581