Calculate the pH after 0.14 mole of NaOH is added to 1.07 L of a solution that is 0.54 M HF and 1.15 M NaF, and calculate the pH after 0.28 mole of HCl is added to 1.07 L of the same solution of HF and NaF.
0.14 mole of NaOH
0.28 mole of HCl
To calculate the pH after adding NaOH and HCl to the solution, we need to consider the reactions that occur and the equilibrium conditions.
First, let's look at the reaction between NaOH and HF:
NaOH + HF --> NaF + H2O
Since NaOH is a strong base and HF is a weak acid, the reaction will form the NaF salt and water, and the HF will be neutralized.
When 0.14 moles of NaOH is added to the solution, it reacts completely with HF. Therefore, the moles of HF remaining will be the initial moles of HF minus the moles of NaOH added:
moles of HF remaining = initial moles of HF - moles of NaOH added
= 1.07 L * 0.54 M - 0.14 mol
= 0.579 - 0.14
= 0.439 mol
Now, let's calculate the pH after adding 0.14 mole of NaOH:
Since HF is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log([A-]/[HA])
In this case, HF is the weak acid (HA) and NaF is the weak conjugate base (A-). The pKa value of HF is 3.17.
pH = 3.17 + log([NaF]/[HF])
= 3.17 + log(1.15 M / 0.439 M)
= 3.17 + log(2.62)
≈ 3.17 + 0.42
≈ 3.59
Therefore, after adding 0.14 mole of NaOH, the pH of the solution would be approximately 3.59.
Now, let's move on to the reaction between HCl and the remaining HF:
HCl + HF --> H2F+ + Cl-
Since HCl is a strong acid and HF is a weak acid, the reaction will form H2F+ (the conjugate acid of HF) and Cl- (the conjugate base of HCl). The HF will be completely neutralized.
When 0.28 moles of HCl is added to the solution, it reacts completely with the remaining HF. Therefore, the moles of HF remaining will be zero.
Since there is no HF left, the Henderson-Hasselbalch equation cannot be used to calculate the pH. Instead, we need to calculate the pH directly using the H3O+ concentration.
The reaction between HCl and water can be written as:
HCl + H2O --> H3O+ + Cl-
Since HCl is a strong acid, it will completely dissociate in water to form H3O+ ions. Therefore, the concentration of H3O+ ions will be equal to the initial concentration of HCl added.
The moles of HCl added can be calculated as follows:
moles of HCl added = 0.28 mol
The concentration of H3O+ ions can be calculated using the formula:
[H3O+] = moles of H3O+ ions / volume of solution
[H3O+] = 0.28 mol / 1.07 L
= 0.261 M
The pH can be calculated using the formula:
pH = -log[H3O+]
pH = -log(0.261)
≈ -(-0.584)
≈ 0.584
Therefore, after adding 0.28 mole of HCl, the pH of the solution would be approximately 0.584.