Consider the titration of 25.0 mL of 0.060 M HCN with 0.065 M NaOH. What is the
ID: 528335 • Letter: C
Question
Consider the titration of 25.0 mL of 0.060 M HCN with 0.065 M NaOH.
What is the pH before any NaOH is added?
What is the pH at the halfway point of the titration?
What is the pH when 95% of the NaOH required to reach the equivalence point has been added?
What is the pH at the equivalence point?
pH initial =
pH at halfway =
pH when 95% is added =
pH at the equivalence point =
Consider the titration of 25.0 mL of 0.060 M HCN with 0.065 M NaOH.
What is the pH before any NaOH is added?
What is the pH at the halfway point of the titration?
What is the pH when 95% of the NaOH required to reach the equivalence point has been added?
What is the pH at the equivalence point?
pH initial =
pH at halfway =
pH when 95% is added =
pH at the equivalence point =
Explanation / Answer
before anything is added
HCN <-> H+ + CN-
Ka = [H*][CN-]/[HCN]
Ka = 6.2×10–10
6.2*10^-10 = x*x/(0.06-x)
x = 6.098*10^-6
pH = -log(H) = -log(6.098*10^-6) = 5.2148
b)
half pint --> pH = pKa + log(A-/HA)
A- = HA in the middle point
so
pH = pKa + log(1)
pKa = -log(6.2*10^-10) = 9.21
pH = pKa
pH = 9.21
c)
95% before
pH = pKa ++ log(A-/HA)
5% is acid, 95% is A- s
pH = 9.21 + log(95/5) = 10.48
d)
pH in equivalence point
Kb = Kw/Ka = (10^-14)/(6.2*10^-10) = 1.61*10^-5
Kb = [OH-][CN-]/[HCN]
[CN-] = M1*V1/(V1+V2) = 25*0.06 / ( 25 + 0.065/(25*0.06))= 0.0598
1.61*10^-5 = x*x/(0.0598-x)
x = OH- = 9.73*10^-4
pH = 14 + log(9.73*10^-4) = 10.98