Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.140M HClO(aq) with 0.140M KOH(aq). The ionization constant for HClO is 4.0*10^-8 which was obtained from this site a) before addition of any KOH? b)after addition of 25.0 mL of KOH? c)after addition of 35.0 mL of KOH? d)after addition of 50.0 mL of KOH? e)after addition of 60.0 mL of KOH ?

Answer 1

pH = 1/2 (pKa - log C)

pH = 1/2 ( -log(4 * 10^-8) - log (50 * 0.14/1000))

pH = 4.776

after addition 25 mLof KOH,

pH = 7 + 1/2 (pKa +log C)

pH = 7 + 1/2 ( -log(4*10^-8) + log ((25*0.14)/(25+50))

pH = 10.033

after addition of 35 mL KOH,

pH = 7 + 1/2 (pKa +log C)

pH = 7 + 1/2 ( -log(4*10^-8) + log ((35*0.14)/(35+50))

pH = 10.08

after addition of 50 mL KOH,

pH = 7 + 1/2 (pKa +log C)

pH = 7 + 1/2 ( -log(4*10^-8) + log (50*0.14/100))

pH = 10.12

after addition of 60 mL KOH,

Number of moles of base remaining = ((60*0.14/1000) - (50*0.14/1000)))

                                                                   = 0.0014

pOH = -log(0.0014) = 2.852

pH = 14 - 2.854 = 11.146