Consider the titration of 20.0 mL of 0.182-M of KX with 0.131-M HCl. The pK a of
ID: 907999 • Letter: C
Question
Consider the titration of 20.0 mL of 0.182-M of KX with 0.131-M HCl. The pKa of HX = 10.86. Give all pH values to 0.01 pH units.
a) What is the pH of the original solution before addition of any acid?
pH = 12.06
b) How many mL of acid are required to reach the equivalence point?
VA = 27.79 mL
c) What is the pH at the equivalence point?
pH = 5.99
d) What is the pH of the solution after the addition of 11.4 mL of acid?
pH = ?
e) What is the pH of the solution after the addition of 33.4 mL of acid?
pH = ?
Explanation / Answer
Consider the titration of 20.0 mL of 0.182-M of KX with 0.131-M HCl. The pKa of HX = 10.86. Give all pH values to 0.01 pH units.
a) What is the pH of the original solution before addition of any acid?
pH = 12.06
Solution :- Using the pka lets calculate the pkb and kb
Pkb = 14 – pka
Pkb = 14 – 10.86 = 3.14
Kb = antilog [-pkb]
Kb= antilog [-3.14]
Kb= 7.24*10^-4
Now using the kb lets calculate the concentration of the OH-
Kb= [HX][OH-]/[X-]
7.24*10^-4 = [x][x]/[0.182-x]
7.24*10^-4 * 0.182-x= x^2
Solving for x we get
0.0111 =x
Therefore
Now lets calculate the pOH
pOH= -log [OH-]
pOH= -log [0.0111]
pOH = 1.95
pH= 14 – pOH
pH= 14 – 1.95
pH = 12.05
b) How many mL of acid are required to reach the equivalence point?
VA = 27.79 mL
Solution :-
Volume of the HCl needed to reach the equivalence point is calculated as
Volume of HCl = molarity of the KX* volume of KX / molarity of HCl
= 0.182 M * 20 ml / 0.131 M
= 27.79 ml
c) What is the pH at the equivalence point?
pH = 5.99
At the equivalence point all the X- is converted to the HX
So the molarity of the HX = 0.182 M * 20 ml / (20+27.79 ml) = 0.0762 M
Now using this concentration lets calculate the H3O+
Ka = [H3O+][X-]/[HX]
Ka = antilog [-pka]
Ka= antilog [-10.86]
Ka= 1.38*10^-11
Lets put the values in the formula
1.38*10^-11 = [x][x]/[0.0762]
1.38*10^-11 *0.0762 = x^2
1.05*10^-12 =x^2
Taking square root of both side we get
1.02*10^-6 = x [H3O+]
pH= -log [H3O+]
pH= -log [1.02*10^-6 ]
pH= 5.99.
d) What is the pH of the solution after the addition of 11.4 mL of acid?
pH = ?
Solution :-
Moles of KX= 0.182 mol per L * 0.020 L = 0.00364 mol
Moles of HCl = 0.131 mol per L * 0.0114 L = 0.0014934 mol
So after the reaction moles of KX remain = 0.00364 – 0.0014934 = 0.0021466 mol
Moles of HX formed = 0.0014934 mol
Total volume = 20 ml + 11.4 ml = 31.4 ml = 0.0314 L
New molarity of the [x-] = 0.0021466 mol / 0.0314 L = 0.06836 M
New molarity of the [HX] =0.0014934 mol / 0.0314 L = 0.04756 M
Using the Henderson equation lets calculate the pH
pH= pka + log [base /acid]
pH= 10.86+ log [0.06836/0.04756]
pH= 11.02
e) What is the pH of the solution after the addition of 33.4 mL of acid?
pH = ?
Solution :-
Moles of KX = 0.182 mole per L * 0.020 L = 0.00364 mol
Moles of HCl = 0.131 mol per L * 0.0334 L = 0.004375 mol
Moles of HCl remain after reaction = 0.004375 mol – 0.00364 mol = 0.000735 mol
Total volume = 20 ml + 33.4 ml = 53.4 ml = 0.0534 L
New molarity of the HCl = 0.000735 mol / 0.0534 L = 0.01376 M
pH= -log [H+ ]
pH= -log [0.01376]
pH= 1.86