Question

Calculate the pH for each of the following cases in the titration of of 100.0 mL of 0.180 M HClO(aq) with 0.180 M KOH(aq). The ionization constant for HClO is 4.0x10^-8

A. After 50.0 mL of KOH, pH=?

B. After the addition of 130.0 mL of KOH, pH=?

Answer 1

Answer – We are given, [HClO] = 0.180 M , volume = 100.0 mL , [KOH] = 0.180 M , Ka for HClO = 4.0*10^-8

We need to calculate the [ClO^-] then moles of ClO^-

We know Ka is too small, so we can neglect x in the 0.180-x

4.0*10^-8 * 0.180 = x^2

So, x = 8.48*10^-5 M

A)pH after added 50.0 mL of KOH –

moles of KOH = 0.180 M * 0.050 L = 0.0175 moles

moles of HClO = 0.180 M * 0.100 L = 0.0180 moles

moles of ClO^- = 8.48*10^-5 M*0.10 = 8.48*10^-6

when we added KOH the moles of acid decrease and its conjugate base increase

moles of HClO = 0.0180 moles – 0.0175 mole = 0.0005 moles

moles of ClO^- = 8.48*10^-6+ 0.0175 moles = 0.0175 moles

total volume = 100 +50 = 150 mL = 0.150 L

so, [HClO] = 0.0005 moles / 0.150 L = 0.00333 M

[ClO^-] = 0.0175 mol / 0.150 L = 0.117 M

pH = pKa + log [ClO^-] / [HClO]

      = 7.40 + log 0.117 / 0.00333

      = 8.94

B) pH after 130.0 mL of 0.180 M KOH

moles of KOH = 0.180 M * 0.130 L = 0.0234 moles

when we added KOH the all moles of acid converted to its conjugate base increase

moles of ClO^- = 8.48*10^-6+ 0.0234 moles = 0.0234 moles

total volume = 100 +130 = 230 mL = 0.230 L

[ClO^-] = 0.0234 mol / 0.230 L = 0.102 M

We need to use the ICE

    ClO^- + H₂O <----> HClO + OH^-

I    0.102                0             0

C     -x              +x              +x

E   0.102-x        +x           +x

We know , Ka = 4.0*10^-8

So, Kb = 1.0*10^-14/ 4.0*10^-8 = 2.5*10^-7

2.5*10^-7 = x*x /0.102-x

We can neglect x in the 0.102-x, since Kb value is too small

So, x² =   2.7*10^-7 *0.102

     x = 1.6*10^-4 M

, so, [OH^-] = 1.6*10^-4 M

pOH = -log [OH^-]

      = - log 1.6*10^-4 M

    = 3.80

So, pH = 14 – pOH

             = 14-3.80

             = 10.2