Question

A 25.0 mL sample of 0.150M HClO is titrated with a 0.150M NaOH. What is the pH at the equilvalence point? Do the molarities matter?

Answer 1

Ka of KClO = 3*10^-8
Initially,number of moles of HCLO = M*V = 0.15M *0.025 L = 0.00375 mol

At equivalence point equal number of NaOH is required.
So, number of moles of NaOH required = 0.00375 mol
Volume of NaoH added = number of moles/M = 0.00375/0.15= 0.025 L= 25 mL
So, 25 mL of NaoH is required

during titration, reaction taking place is:
                HClO +     NaOH ----> ClONa + H2O
initial:     0.00375        0.0375        0             0
final:        0                0            0.00375     0.0375
So, number of moles of CloNa formed = 0.00375 moles
and total volume is 50 ml = 0.05 L
[CloNa] = n/V = 0.00375/0.05 = 0.075 M

This CloNa get hydrolysed as:
CloNa + H2O ----> HClO +OH -
0.075                             0           0
0.075-x                        x             x
Kb of CloNa = Kw/Ka = 10^-14 / ( 3*10^-8)
                 =3.33*10^-7
Kb = x*x / (0.075-x)
since Kb is very small x can be ignored in comparison to 0.075
so,
Kb = x*x / (0.075)
3.33*10^-7 = x^2/0.075
x=1.58*10^-4
so, [OH-] = 1.58*10^-4
pOH = -log [OH-] = -log (1.58*10^-4) =3.8
pH= 14- pOH = 14- 3.8 = 10.2

Ph will be 3.8

Molarimeter accuracy will surely matter because whole calculation is based on number of moles