The pH of a solution that contains 0.818 M acetic acid (Ka = 1.76 x 10-5) and 0.172 M sodium acetate is
__________. The Ka of acetic acid is 1.76 × 10-5.
4.07
To find the pH of the solution, we first need to determine the concentration of the acetate ion (CH3COO-) in the solution.
Acetic acid (CH3COOH) is a weak acid, and it partially dissociates in water. The dissociation reaction can be written as follows:
CH3COOH ⇌ CH3COO- + H+
From the given information, we know that the concentration of acetic acid is 0.818 M. Since acetic acid is a weak acid, we can assume that the initial concentration of CH3COO- is negligible. Therefore, the concentration of CH3COO- can be considered equal to the initial concentration of sodium acetate (NaCH3COO), which is 0.172 M.
Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution, which is given by:
pH = pKa + log([CH3COO-]/[CH3COOH])
Here, pKa is the negative logarithm of the acid dissociation constant (Ka), which is -log(1.76 x 10-5).
Let's substitute the values:
pH = -log(1.76 x 10-5) + log(0.172/0.818)
Now, we can calculate the pH using a calculator or software:
pH ≈ 4.77
Therefore, the pH of the solution is approximately 4.77.
To find the pH of a solution containing both acetic acid and sodium acetate, we need to consider the ionization of acetic acid and the common ion effect.
The ionization of acetic acid can be represented by the following equation:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
Given that the Ka of acetic acid is 1.76 x 10^-5, we can set up an expression for the equilibrium constant (Keq):
Ka = [CH3COO-][H+]/[CH3COOH]
Since the solution also contains sodium acetate (CH3COONa), which dissociates completely into CH3COO- and Na+, the concentration of CH3COO- is determined by the concentration of sodium acetate.
To calculate the pH, we need to determine the amount of acetic acid that has ionized and the concentration of H+. We can consider the initial concentrations of acetic acid and sodium acetate as the total concentrations.
First, we need to calculate the concentration of CH3COO- by multiplying the concentration of sodium acetate (0.172 M) by the ratio of CH3COO- to sodium acetate in the compound. Since one mole of sodium acetate generates one mole of CH3COO-, the concentration of CH3COO- is also 0.172 M.
Now, let x represent the concentration of H+ that has been formed from the dissociation of acetic acid. Therefore, the concentration of CH3COOH that has ionized is (0.818 - x) M.
Using the equilibrium constant expression:
Ka = (0.172 M)(x)/(0.818 - x)
Rearranging the equation to solve for x:
x = Ka * (0.818 - x)/(0.172)
Now we can substitute the value of Ka (1.76 x 10^-5) and solve for x.
x = (1.76 x 10^-5) * (0.818 - x)/(0.172)
Simplifying this equation will give us a quadratic equation, which we can solve using various methods, such as factoring or the quadratic formula.
After solving for x, we can calculate the concentration of H+. The pH can be determined using the equation pH = -log[H+].