(a) When 0.10 mol of the ionic solid NaX, where X is an unknown anion, is dissol
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Question
(a) When 0.10 mol of the ionic solid NaX, where X is an unknown anion, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 8.66. When 0.10 mol of the ionic solid ACl, where A is an unknown cation, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 7.00. What would be the pH of 1.0 L of solution that contained 0.10 mol of AX? Be sure to document how you arrived at your answer. (Do this on paper. Your instructor may ask you to turn in this work.)
(c) From the information presented in part (a), calculate Kb for the X (aq) anion.
From the information presented in part (a), calculate Ka for the conjugate acid of X (aq).
(g) Some students mistakenly think that a solution that contains 0.10 mol of AX and 0.10 mol of HCl should have a pH of 1.00. Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong. (Select all that apply.)
-Since the conjugate acid is a weak acid, you would not expect it to dissociate completely, so the pH would not be 1.00.
-Students have the misconception that the weak base dominates the behavior, but this is not the case.
-This is a solution of equal moles of a strong acid and a weak acid.
-This is a solution of equal moles of a strong acid and a weak base.
-When estimating the pH, you count the position of the first non-zero decimal, which approximates the pH.
-The HCl will consume all the AX, which would result in a neutral solution with a pH of 7.00.
-Students have the misconception that the strong acid dominates the behavior, but this is not the case.
-The acid and the base will react with each other and produce an equal number of moles of conjugate acid HX.
Explanation / Answer
a)
With NaX
X- + H2O --> HX + OH-
0.1 – a --> a + a
pOH = 14 – pH = 14 – 8.66 = 5.34
[OH-] = a = 10-5.34 = 4.57 x 10-6 M
Kb of X- = a2 / (0.1 – a) = 2.09 x 10-10
With ACl
pH = 7
This implies that neither A+ nor Cl- will undergo hydrolysis with water molecules
Kb of Cl- = 10-14/ 0.1 = 10-13
Ka of HCl = Kw/ Kb = 0.1
b)
pH of AX will same as pH of NaX = 8.66
c)
Kb of X- = 2.09 x 10-10
d)
Ka of HX = Kw/Kb = 10-14 / 2.09 x 10-10
Ka = 4.79 x 10-5
d)
Take Ka of HCl = 0.1 as calculated in part A
X- + H2O --> HX + OH-
0.1 – y --> y + y
HCl --> H+ + Cl-
0.1 - z --> z + z
Ka of HCl = z2 / (0.1 – z) = 0.1
z = 0.0618
pH = -log(z) 1.2 1
Weak conjugate base and strong acid will interact with each other.
The options that apply are:
1) Statement 1 is not true as strong acid will dominate over weak conjugate acid.
2) Weak base will also not dominate.
3) This is a solution of equal moles of a strong acid and a weak base.
4) The first non-zero decimal is at 1/ 100th place i.e. pH is between 1 and 2.
5) The HCl will not consume all the AX. AX will be in equilibrium as per Ka and Kb values.
6) Although strong acid will dominate in determination of pH but we cannot totally neglect the contribution of weak base in H+ consumption.
7) The acid and the base will react with each other, but will not produce an equal number of moles of conjugate acid HX. HX will be in equilibrium as per its Ka value.
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