pH of 10^-3M HAc and 10^-2M NH4Cl in aqueous solution?? Do you just figure out each [H+] concentration separately and then add them??
To determine the pH of a solution containing both an acid and its conjugate base, you need to consider the principles of acid-base chemistry and the equation for the equilibrium constant (Ka) of the acid.
1. Start by writing the balanced equation for the dissociation of the acid (HAc):
HAc ⇌ H+ + Ac-
2. The equilibrium constant (Ka) for this reaction is given by:
Ka = [H+][Ac-] / [HAc]
3. Since the concentration of the acid (HAc) is given as 10^-3 M, the initial concentration of H+ is also 10^-3 M.
4. To find the concentration of the conjugate base (Ac-) at equilibrium, you need to know the value of Ka (acid dissociation constant). For acetic acid (HAc), the Ka value is approximately 1.8 x 10^-5.
5. Now, apply the equation for Ka and the equilibrium concentrations to find the concentration of Ac-:
1.8 x 10^-5 = (10^-3)([Ac-] / 10^-3)
[Ac-] = 1.8 x 10^-5 M
6. Next, consider the dissociation of NH4Cl in water:
NH4Cl ⇌ NH4+ + Cl-
7. The ammonium ion (NH4+) acts as a weak acid, and its conjugate base (Cl-) is the strong base. This means that NH4Cl will produce an acidic solution when dissolved in water.
8. Since the concentration of NH4Cl is given as 10^-2 M, the concentration of NH4+ is also 10^-2 M.
9. Calculate the concentration of OH- (hydroxide ions) in the solution by applying the Kw (water ion product) constant value of 1.0 x 10^-14.
Kw = [H+][OH-] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / 10^-3
[OH-] = 1.0 x 10^-11 M
10. Since water is neutral, the concentration of H+ is equal to the concentration of OH-. Therefore, the concentration of H+ in the solution is also 1.0 x 10^-11 M.
11. Finally, calculate the pH of the solution by taking the negative logarithm (base 10) of the H+ concentration:
pH = -log[H+]
pH = -log(1.0 x 10^-11)
pH ≈ 11
So, in the given solution, the pH would be approximately 11.