pH and Tritation Problems 23 assignment help
1. Calculate the pH of a solution that is 0.1 M in acetic acid (Ka = 1.8 x 10^-5).
To calculate the pH of the solution, we will first write out the dissociation reaction of acetic acid:
CH3COOH + H2O -> CH3COO- + H3O+
Next, we need to set up an ICE table to track the concentrations of the species at equilibrium:
CH3COOH + H2O -> CH3COO- + H3O+
I 0.1 M 0 M 0 M
C -x +x +x
E 0.1-x x x
Using the equilibrium constant expression for Ka, we can write:
Ka = [CH3COO-][H3O+]/[CH3COOH]
1.8 x 10^-5 = x^2 / (0.1 - x)
Since the value of x is expected to be very small compared to 0.1, we can approximate (0.1 - x) to be 0.1:
1.8 x 10^-5 = x^2 / 0.1
x = sqrt(0.1 * 1.8 x 10^-5) = 4.24 x 10^-3
Therefore, the concentration of H3O+ is 4.24 x 10^-3 M. To find the pH, we take the negative logarithm of the H3O+ concentration:
pH = -log(4.24 x 10^-3) = 2.37
Therefore, the pH of the 0.1 M acetic acid solution is 2.37.
2. A solution of hydrochloric acid (HCl) has a pH of 1.8. What is the concentration of HCl in the solution?
Given that the pH of the solution is 1.8, we can find the concentration of H3O+ using the equation:
pH = -log[H3O+]
1.8 = -log[H3O+]
[H3O+] = 10^(-1.8) = 1.58 x 10^-2 M
Since HCl dissociates completely in water to produce H3O+, the concentration of HCl in the solution is equal to the concentration of H3O+:
[HCl] = 1.58 x 10^-2 M
Therefore, the concentration of HCl in the solution is 1.58 x 10^-2 M.