Question

what is the pH when titrating 100 mL of 0.180M HClO with 100mL of KOH, also afer 130 mL of KOH is added. Ka=4x10^-8

Answer 1

HClO + KOH -----> KClO + H2O

MOles of HClO = Molarity*V in L = 0.18*100*10^-3 = 0.018

Moles of KOH = Molarity*V = 0.18*100*10^-3 = 0.018 [Molarity is not given taken as same as HClO.i.e 0.18 M ]

HClO + KOH ----> KClO + H2O

0.018 0.018 0

0 0 0.018

[KClO] = 0.018 / Total volume in L = 0.018 / 200*10^-3 = 0.09

ClO- + H2O -----> HClO + OH-

0.09 0 0

0.09-x x x

Kb = x^2 / 0.09-x = Kw/Ka = 10^-14 / 4*10^-8 = 2.5*10^-7

2.5*10^-7 = x^2 / 0.09-x

x^2 + 2.5*10^-7 x - 2.25*10^-8 = 0

x = 1.498*10^-4

pOH = -log x = 3.82

pH = 10.18

If 130 ml of KOH is added of molarity 0.18M

Moles of KOH = 0.0234

KOH + HClO -----> KClO + H2O

0.0234 0.018 0

0.0234-0.018 0 0.018

[OH-] = 0.0234-0.018 / 230*10^-3 = 0.0234

pOH = 1.63

pH = 12.37