1. In a problem involving the mixing of a weak acid solution with a strong base
ID: 530457 • Letter: 1
Question
1. In a problem involving the mixing of a weak acid solution with a strong base solution (such as the titration of a weak acid with a strong base), before attempting to calculate the pH of the resulting solution, you must ___.
A.) Determine which (the weak acid or strong base) is present in excess
B.) Determine the pH of the weak acid solution.
C.) determine th pH of the strong base solution
D.) Determine the difference in volume of the acid and base solutions.
2.) Consider a weak acid , HA, for which the pKa is 5.826. The pH of a 0.28 M solution of this weak is is ___ at 25 oC. (Answer to three decimal places.)
3.) What is the pH of the solution resulting from mixing 33.07 mL of 0.183 M HA (pKa = 4.876) with 14.82 mL of 0.100 M KOH? (answer to three decimal places)
pH = ___
4.)
What is the pH of a solution created by mixing 32.19 mL of 0.200 M HA (Ka = 1.81 x 10-5 M) with 32.19 mL of 0.200 M KOH? (Answer to three decimal places)
pH = ___
5.)
What is the pH of a solution created by mixing 18.52 mL of 0.129 M HA (pKa = 4.877) with 73.61 mL of 0.100 M KOH? (Answer to three decimal places)
pH = ___
A.) Determine which (the weak acid or strong base) is present in excess
B.) Determine the pH of the weak acid solution.
C.) determine th pH of the strong base solution
D.) Determine the difference in volume of the acid and base solutions.
Explanation / Answer
1) We should know that which (weak acid or strong base) is present in excess
Answer : A
2.) Consider a weak acid , HA, for which the pKa is 5.826. The pH of a 0.28 M solution of this weak is is ___ at 25 oC. (Answer to three decimal places.)
pKa = 5.826 = -logKa
Ka = 1.49 X 10^-6
[H+] = (Ka X concentration of weak acid)1/2
[H+] = (1.49 X 10-6 X 0.28)1/2 = 6.46 X 10^-4
pH = -log[H+] = 3.189
3.) What is the pH of the solution resulting from mixing 33.07 mL of 0.183 M HA (pKa = 4.876) with 14.82 mL of 0.100 M KOH? (answer to three decimal places)
Answer: the moles of acid = Molarity X volume = 0.183 x 33.07 / 1000 = 6.052 millimoles
Moles of base added = Molarity x volume = 0.1 x 14.82 / 1000 = 1.482 millimoles
These millimoles of base will react with 1.482 millimoles of acid to form 1.482 millimoles of salt
The millimoles of acid left = 6.052 - 1.482 = 4.57 millimoles
this will form a buffer containing weak acid and its salt with strong base
pH = pKa + log [salt] / [acid]
pH = 4.875 + log [1.482] /[4.57] = 4.386
4.)
What is the pH of a solution created by mixing 32.19 mL of 0.200 M HA (Ka = 1.81 x 10-5 M) with 32.19 mL of 0.200 M KOH? (Answer to three decimal places)
Moles of acid present = molarity X volume = 0.2 X 32.19 / 1000 = 6.438 millimoles
Moles of base added = molarity X volume = 0.2 X 32.19 / 1000 = 6.438 millimoles
so there will be complete neutralization
The moles of salt formed = 6.438 millimoles
[Salt ] =Millimoles/ total volume in mL = 6.438 / 32.19 + 32.19 = 0.1 M
The salt will hydrolyze as
A- + H2O ---> AH + OH-
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x x x
Kb = Kw / Ka = 10^-14 / 1.81 X 10^-5 = [AH][OH-]/[A-] = 5.525 X 10^-10
5.525 X 10^-10 = x^2 / (0.1-x) ,we can ignore x in denominator as x <<< 0.1
x = 7.433 X 10^-6 M = [OH-]
pOH = -log [OH-] = 5.129
pH = 8.871
5.)
What is the pH of a solution created by mixing 18.52 mL of 0.129 M HA (pKa = 4.877) with 73.61 mL of 0.100 M KOH? (Answer to three decimal places)
The moles of acid added = molarity X volume = 0.129 x 18.52 / 1000 = 2.389 millimoles
Moles of KOH added = Molarity X volume = 0.1 x 73.61 / 1000 = 7.361 millimoles
Millimoles of KOH reacted = 2.389 millimoles
Millimoles of KOH left = 7.361 - 2.389 = 4.972
[OH-] after neutralization = millimoles of KOH left / Total volume in mL = 4.972 / 18.52 + 73.61 = 0.054 M
pOH = -logOH- ] = 1.268
pH = 12.732
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