Question

a 50.0 mL sample of a 0.100 M solution of NaCN is titrated by 0.100 M HCl. kb for CN is 2.0x10-5. A.calculate the pH of the solution prior to the start of the titration. B after the addition of 10.0 mL. C. after the addition of 25.0 mL of 0.100 M HCl. D. at the equivalence point. E. after the addition of 60.0 mL of 0.100 M HCl

Answer 1

pOH = 1/2 (pKb - log C)
pOH = 1/2(-log(2*10^-5) - log(0.1*50))
pOH = 2
pH = 14 - 2 = 12
pH = pKb + log(salt)/(base)
pH = -log(2 * 10^-5) + log((0.1*10)/(0.1*50))
pH = 4
pH = pKb + log(salt)/(base)
pH = -log(2 * 10^-5) + log((0.1*25)/(0.1*50))
pH = 4.398
at equivalence point number of moles of acid = number of moles of base
pH = 7 - 1/2(pKb + logC)
pH = 7 - 1/2(-log(2*10^-5) + log(0.1*50/(50+50)))
pH = 5.301
number of moles of HCl remainig = 60*0.1 - 50*0.1 = 1 mmol
pH = -log(H+) = -log(1*10^-3)
pH = 3