Consider the diprotic acid H2A with K1 1*10^-3 and K2= 1*10^-9 .
Find pH, [H2A], [HA], and [A2-] for a .2000M solution of H2A
b)Find pH, [H2A], [HA], and [A2-] for a .200M solution of HA-
c) Find pH, [H2A], [HA], and [A2-] for a .200M solution of A-
To solve this problem, we can use the principles of acid-base equilibrium and the equations derived from the Law of Mass Action. We will calculate the pH, as well as the concentrations of the different species H2A, HA, and A2- for each scenario.
a) For a .2000M solution of H2A:
- The initial concentration of H2A is .2000M.
- Since H2A is a diprotic acid, it can donate two protons (H+ ions) in solution. The equilibrium reactions are as follows:
H2A ⇌ H+ + HA (Equation 1)
HA ⇌ H+ + A2- (Equation 2)
- From Equation 1, the equilibrium constant K1 is given as 1*10^-3, which represents the ratio of [H+][HA] to [H2A]. We can use this expression to calculate [H+][HA].
- Similarly, from Equation 2, the equilibrium constant K2 is given as 1*10^-9, which represents the ratio of [H+][A2-] to [HA]. We can use this expression to calculate [H+][A2-].
Now let's calculate the pH and concentrations:
Step 1: Calculate [H+][HA] using K1
K1 = [H+][HA] / [H2A]
1*10^-3 = [H+][HA] / .2000M (Note: [H2A] = .2000M)
[H+][HA] = (1*10^-3) * .2000M
Step 2: Calculate [H+][A2-] using K2
K2 = [H+][A2-] / [HA]
1*10^-9 = [H+][A2-] / [HA]
[H+][A2-] = (1*10^-9) * [HA]
Step 3: Create the charge balance equation
[H+][HA] + [H+][A2-] = [OH-] + [H2A] + [HA]
Since the solution contains no hydroxide ions (OH-), [OH-] can be ignored.
Step 4: Use the mass balance equation to relate the concentrations of H2A, HA, and A2-
[H2A] = .2000M - [H+]
[HA] = [H+][HA] / K1 = ([H+][HA] / (.2000M)) / K1
[A2-] = [H+][A2-] / K2 = ([H+][A2-] / [HA]) / K2
Step 5: Calculate the pH
pH = -log10[H+]
Now, you can substitute the calculated values into the equations to get the final results for the concentrations and pH.
b) For a .200M solution of HA-:
- In this case, the initial concentration of HA is .200M, which is the same as [HA].
- Since HA- acts as a weak acid, it can accept a proton (H+) to form H2A. The equilibrium reaction is as follows:
HA ⇌ H+ + A2- (Equation 2)
- We will use Equation 2 (mentioned in part a) to calculate the pH and concentrations.
- The steps are the same as in part a), but in this case, [HA] is already known as .200M.
c) For a .200M solution of A-:
- In this case, the initial concentration of A2- is .200M, which is the same as [A2-].
- A2- acts as a strong base and cannot accept any more protons (H+).
- The pH of a strong base solution depends on the concentration of the hydroxide ion (OH-), which is determined by the concentration of A2-.
- We can use the equation [OH-] = Kw / [A2-] to calculate [OH-].
- Since [OH-] is known, we can calculate the pOH and then the pH using the formula pH = 14 - pOH.