What is the pH of 1.00 L of the buffer solution that contains 0.100 M HF and 0.120 M NaF after 0.020 mol of HNO3 is added? The Ka of HF is 3.5×10−4, and so pKa = 3.46.
To determine the pH of the buffer solution after adding HNO3, we need to consider the reaction that occurs between HNO3 and the components of the buffer.
The reaction between HNO3 and HF is: HNO3 + HF → H2O + NO3- + F-
Since HNO3 is a strong acid and completely dissociates, we can assume that after the reaction, we have 0.020 mol of NO3- and F- ions in our solution.
Now, let's calculate the concentrations of HF and NaF after the reaction:
Initially, we have 0.100 M HF and 0.120 M NaF. Since NaF is a strong electrolyte and completely dissociates, we have 0.120 M F- ions.
After the reaction, the initial moles of HF decrease by 0.020 mol, but the volume of the solution remains the same. Therefore, the final concentration of HF is (0.100 mol - 0.020 mol) / 1.00 L = 0.080 M.
Now we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([A-] / [HA])
Where [A-] is the concentration of the conjugate base (F-) and [HA] is the concentration of the acid (HF).
pH = 3.46 + log (0.120 M / 0.080 M)
Calculating this expression, we find:
pH = 3.46 + log (1.5)
Using logarithmic properties, the equation simplifies to:
pH ≈ 3.46 + 0.176
Calculating this, we get:
pH ≈ 3.636
Therefore, the pH of the buffer solution after adding 0.020 mol of HNO3 is approximately 3.636.