Part A:
To find the pH of a 0.130 M solution of a weak monoprotic acid with Ka = 1.0×10-5, we can use the following equation:
Ka = [H+][A-]/[HA]
Where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid. Since the acid is monoprotic, the concentration of the acid is the same as the concentration of the initial solution.
We can set up an ICE table to find the concentration of each species:
HA + H20 ↔ H3O+ + A-
↔ H3O+ + A-I 0.130 M 0 M 0 M
↔ H3O+ + A-I 0.130 M 0 M 0 MC -x +x +x
↔ H3O+ + A-I 0.130 M 0 M 0 MC -x +x +xE 0.130-x x x
Substituting these values into the Ka expression, we get:
1.0×10^-5 = (x^2)/(0.130-x)
Solving for x using the quadratic formula, we get x = 3.162×10^-3 M. This is the concentration of [H+].
Taking the negative log of [H+], we get the pH:
pH = -log[H+] = -log(3.162×10^-3) = 2.50
Therefore, the pH of a 0.130 M solution of a weak monoprotic acid with Ka = 1.0×10-5 is 2.50
Part B:
To find the percent ionization of the weak acid, we can use the equation:
% Ionization = ([H+]/[HA]) × 100
% Ionization = ([H+]/[HA]) × 100From Part A, we found that [H+] = 3.162×10^-3 M and [HA] = 0.130 M, so:
% Ionization = ([H+]/[HA]) × 100From Part A, we found that [H+] = 3.162×10^-3 M and [HA] = 0.130 M, so:% Ionization = (3.162×10^-3/0.130) × 100 = 2.43%
Therefore, the percent ionization of a 0.130 M solution of a weak monoprotic acid with Ka = 1.0×10−5 is 2.43%.
Part C:
To find the pH of a 0.130 M solution of a weak monoprotic acid with Ka = 1.3×10^-, we can use the same method as in Part A.
Setting up an ICE table:
HA + H2O ↔ H3O+ + A-
HA + H2O ↔ H3O+ + A-I 0.130 M 0 M 0 M
HA + H2O ↔ H3O+ + A-I 0.130 M 0 M 0 MC -x +x +x
HA + H2O ↔ H3O+ + A-I 0.130 M 0 M 0 MC -x +x +xE 0.130-x x x
Substituting these values into the Ka expression, we get:
1.3×10^-3 = (x^2)/(0.130-x)
Solving for x, we get x = 0.0361 M. This is the concentration of [H+].
Taking the negative log o [H+], we get the pH:
pH:pH = -log[H+] = -log(0.0361) = 1.44
Therefore, the pH of a 0.130 M solution of a weak monoprotic acid with Ka = 1.3×10^-3 is 1.44.
Ka = 1.3×10^-3 is 1.44.Part D:
To find the percent ionization of the weak acid, we can use the same equation as in Part B:
% Ionization = ([H+]/[HA]) × 100
% Ionization = ([H+]/[HA]) × 100From Part C, we found that [H+] = 0.0361 M and [HA] = 0.130 M, so:
% Ionization = ([H+]/[HA]) × 100From Part C, we found that [H+] = 0.0361 M and [HA] = 0.130 M, so:% Ionization = (0.0361/0.130) × 100 = 27.8%
Therefore, the percent ionization of a 0.130 M solution of a weak monoprotic acid with Ka = 1.3×10^-3 is 27.8%.
Part E:
To find the pH of a 0.130 M solution of a weak monoprotic acid with Ka = 0.13, we can use the same method as in Part A
Setting up an ICE table:
HA + H2O ↔ H3O+ + A-
HA + H2O ↔ H3O+ + A-I 0.130 M 0 M 0 M
HA + H2O ↔ H3O+ + A-I 0.130 M 0 M 0 MC -x +x +x
HA + H2O ↔ H3O+ + A-I 0.130 M 0 M 0 MC -x +x +xE 0.130-x x x
Substituting these values into the Ka expression, we get:
0.13 = (x^2)/(0.130-x)
Solving for x, we get x = 0.191 M. This is the concentration of [H+].
Taking the negative log of [H+] we get the pH:
Therefore, the pH of a 0.130 M solution of a weak monoprotic acid with Ka = 0.13 is 0.72.
Part F:
To find the percent ionization of the weak acid, we can use the same equation as in Part B:
% Ionization = ([H+]/[HA]) × 100
% Ionization = ([H+]/[HA]) × 100From Part E, we found that [H+] = 0.191 M and [HA] = 0.130 M, so:
% Ionization = ([H+]/[HA]) × 100From Part E, we found that [H+] = 0.191 M and [HA] = 0.130 M, so:% Ionization = (0.191/0.130) × 100 = 147%
Note that the percent ionization is greater than 100%. This is because the acid is relatively strong (compared to the previous examples) and ionizes to a greater extent.
Note that the percent ionization is greater than 100%. This is because the acid is relatively strong (compared to the previous examples) and ionizes to a greater extent.Therefore, the percent ionization of a 0.130 M solution of a weak monoprotic acid with Ka = 0.13 is 147%.