Consider the titration of 25.0 mL of 0.116-M of KX with 0.144-M HCl. The pK a of
ID: 503640 • Letter: C
Question
Consider the titration of 25.0 mL of 0.116-M of KX with 0.144-M HCl. The pKa of HX = 10.67. Give all pH values to 0.01 pH units.
a) What is the pH of the original solution before addition of any acid?
pH =
b) How many mL of acid are required to reach the equivalence point?
VA =
c) What is the pH at the equivalence point?
pH =
d) What is the pH of the solution after the addition of 11.7 mL of acid?
pH =
e) What is the pH of the solution after the addition of 24.1 mL of acid?
pH =
k -XX"x-)x-* )#7 *Explanation / Answer
pKa of HX = 10.67
a)
It is a salt of strong base and weak acid. so pH > 7
pH = 7 + 1/2 (pKa + log C)
= 7 +1/2 (10.67 + log 0.116)
= 11.87
pH = 11.87
b) at equivalence point :
millimoles of salt = millimoles of acid
25 x 0.116 = 0.144 x V
Volume of acid = 20.1 mL
c)
KX + HCl <--------------> KCl + HX
2.9 2.9 0 0
0 0 2.9 2.9
here only HX is remained.
concentration of [HX] = 2.9 / 25 + 20.1 =0.0643 M
pH = 1/2 (pKa - log C)
= 1/2 (10.67 - log 0.0643)
pH = 5.93
d)
millimoles of acid = 11.7 x 0.144 = 1.68
KX + HCl <--------------> KCl + HX
2.9 1.68 0 0
1.22 0 1.68 1.68
pH = pKa + log [salt / acid]
= 10.67 + log [1.22 / 1.68]
= 10.53
pH = 10.53
e)
millimoles of acid = 24.1 x 0.144 = 3.47
KX + HCl <--------------> KCl + HX
2.9 3.47 0 0
0 0.57 2.9 2.9
[HCl] = 0.57 / (24.1 + 25) = 0.116
pH = -log [H+] = -log (0.116)
pH = 1.94