To calculate the pH of a solution, we need to consider the concentration of H+ ions. For each given solution, we can use the equilibrium constant (Ka or Kb) to determine the concentration of H+ or OH- ions. For example, (a) The pH of 0.05 M HCl is 1.30. (b) The pH of 0.1 M acetic acid is 2.73.
To calculate the pH of a solution, we need to consider the concentration of H+ ions in the solution. We can use the equilibrium constant(Ka or Kb) to determine the concentration of H+ or OH- ions. Let's go through the steps for each solution:
(a) 0.05 M HCl:
HCl is a strong acid and will completely dissociate in water to give H+ ions. Therefore, the concentration of H+ ions will be equal to 0.05 M. Taking the negative logarithm of the concentration, we get a pH of -log(0.05) = 1.30.
(b) 0.1 M acetic acid (CH3COOH):
Since acetic acid is a weak acid, it will partially dissociate in water. The equilibrium constant (Ka) is given as 1.75 × 10-5. To find the concentration of H+, we can set up an ICE (initial, change, equilibrium) table. Let's assume x as the concentration of H+ ions formed from the dissociation of acetic acid. The equilibrium concentration of acetic acid will be (0.1 - x), and the equilibrium concentration of H+ ions will be x. Plugging these values into the Ka expression, we get 1.75 × 10-5 = (x^2) / (0.1 - x). Solving this equation, we find x = 1.868 × 10-3 M. Taking the negative logarithm of this concentration, we get a pH of -log(1.868 × 10-3) = 2.73.
(c) 0.1 M aniline (C6H5NH2):
Aniline is a weak base, and its equilibrium constant (Kb) is given as 3.82 × 10-10. Similar to the previous example, we can set up an ICE table, assuming x as the concentration of OH- ions formed from the dissociation of aniline. The equilibrium concentration of aniline will be (0.1 - x), and the equilibrium concentration of OH- ions will be x. Plugging these values into the Kb expression, we get 3.82 × 10-10 = (x^2) / (0.1 - x). Solving this equation, we find x = 1.933 × 10-6 M. Since we're calculating the pH, we need to find the concentration of H+ ions, which can be obtained by dividing the Kw (1.0 × 10-14) by the concentration of OH-. Therefore, the concentration of H+ ions will be 1.0 × 10-14 / 1.933 × 10-6 = 5.178 × 10-9 M. Taking the negative logarithm of this concentration, we get a pH of -log(5.178 × 10-9) = 8.29.
(d) 0.1 M acetic acid plus 0.001 M HCl:
In this case, we have a mixture of a weak acid (acetic acid) and a strong acid (HCl). The concentration of H+ ions from the strong acid will be equal to 0.001 M. For the weak acid, we can follow the same steps as in part (b) to find the concentration of H+ ions. After finding the concentrations of H+ ions from both acids, we can add them together to get the total concentration of H+ ions. Taking the negative logarithm of this concentration will give us the pH of the solution.
(e) 0.16 M acetic acid plus 0.044 M sodium acetate:
Here, we have a mixture of a weak acid (acetic acid) and its conjugate base (sodium acetate). By comparing the concentrations of the acid and the conjugate base, we can determine which component will have a greater influence on the pH. In this case, the concentration of the acid (0.16 M) is greater than the concentration of the conjugate base (0.044 M). Therefore, we can assume that the solution will be acidic, and the pH will be determined by the concentration of H+ ions from the weak acid. We can follow the same steps as in part (b) to find the concentration of H+ ions and then calculate the pH.