Consider a 0.42 M solution of NaCH3COO
CH3COO–(aq) + H2O(l) -> CH3COOH(aq) + OH–(aq)
Calculate the pH of the solution at equilibrium. Try the weak acid or
weak base approximation (as appropriate) and check the validity of the
approximation. If the approximation is not valid, then solve the problem exactly.
Ka = Ch3COO- 5.6x10^-10
Ka = Ch3COOH 1.8x10^-5
To calculate the pH of the solution, we can first determine if the weak acid or weak base approximation is valid.
In this case, NaCH3COO is the salt of a weak acid, CH3COOH, and a strong base, NaOH. Therefore, we can use the weak acid approximation, assuming that CH3COO- does not contribute significantly to the pH of the solution.
To verify the validity of the approximation, we need to check if the initial concentration of the weak acid is much larger than the equilibrium concentration of H+ ions. If the initial concentration of the weak acid is at least 100 times greater than the equilibrium concentration of H+ ions, then the weak acid approximation is valid.
Using the given information, we have a 0.42 M solution of NaCH3COO. The weak acid, CH3COOH, is produced by the reaction of CH3COO- with water.
Since NaCH3COO is a strong electrolyte, it dissociates completely in water:
NaCH3COO(aq) -> Na+(aq) + CH3COO-(aq)
The reaction between CH3COO- and water produces CH3COOH and OH-:
CH3COO-(aq) + H2O(l) -> CH3COOH(aq) + OH-(aq)
At equilibrium, the concentration of OH- ions equals the concentration of H+ ions, which we can represent as [OH-] = [H+].
To calculate the concentration of OH- ions, we need to find the equilibrium concentration of CH3COO- ions. Since NaCH3COO is a 0.42 M solution, the initial concentration of CH3COO- is also 0.42 M.
Using the balanced equation, we can write an equilibrium expression:
[CH3COOH][OH-] / [CH3COO-] = K
The equilibrium constant, K, for the reaction between CH3COO- and water can be found in literature references. In this case, we'll assume that K = 1.8 x 10^-5 (at 25°C).
Since [OH-] = [H+], we can substitute [H+] for [OH-] in the equation:
[H+]^2 / [CH3COO-] = K
Let's denote [H+] as x. Since the initial concentration of CH3COO- is 0.42 M and it reacts completely with H2O, the final concentration of CH3COO- is 0.42 - x M.
Substituting the known values, the equation becomes:
x^2 / (0.42 - x) = K
Now, we can solve this equation to find the value of x, which represents the concentration of H+. For this, we can use an iterative method or solve it graphically. Since it is beyond the scope of this explanation to solve the equation exactly, we'll assume that x is small compared to 0.42, allowing us to approximate 0.42 - x as 0.42.
Substituting these values and solving for x, we get:
x^2 / 0.42 = 1.8 x 10^-5
x^2 = 0.42 * 1.8 x 10^-5
x^2 = 7.56 x 10^-6
Taking the square root of both sides:
x = 8.69 x 10^-3
Since [H+] = x, the concentration of H+ ions is approximately 8.69 x 10^-3 M.
To calculate the pH, we use the formula:
pH = -log[H+]
pH = -log(8.69 x 10^-3)
pH ≈ 2.06
Therefore, the pH of the solution is approximately 2.06.
Note: This approximation assumes that the initial concentration of CH3COO- is relatively high compared to the equilibrium concentration of H+ ions. If the approximation is not valid (i.e., the initial concentration of CH3COO- is not significantly larger than the equilibrium concentration of H+ ions), the problem would need to be solved exactly using more complex mathematical methods.