Calculate the pH after 0.13 mol of NaOH is added to 1.00 L of the solution that is 0.57 M HF and 1.13 M KF.
I got the answer of 3.66 to 3 sig figs, but it's wrong. Can you please tell me what I did wrong?
-Thanks
To solve this problem, we need to consider the reaction between HF and NaOH and the subsequent changes in pH due to the addition of NaOH.
The balanced equation for the reaction between HF and NaOH is as follows:
HF + NaOH → NaF + H2O
Since NaOH is a strong base and completely dissociates in water, we can assume the stoichiometry of the reaction is 1:1. This means that for every 1 mol of NaOH added, an equal amount of HF will react.
Given that 0.13 mol of NaOH is added to 1.00 L of the solution, we can determine the initial concentration of HF (before the addition) and the change in concentration after the reaction.
Initial moles of HF = Initial concentration of HF x volume of solution
= 0.57 mol/L x 1.00 L
= 0.57 mol
Since the stoichiometry of the reaction is 1:1, the final moles of HF will be equal to the initial moles minus the moles of NaOH added.
Final moles of HF = Initial moles of HF - moles of NaOH added
= 0.57 mol - 0.13 mol
= 0.44 mol
Now, we need to calculate the final concentration of HF after the addition of NaOH.
Final concentration of HF = Final moles of HF / volume of solution
= 0.44 mol / 1.00 L
= 0.44 M
To determine the pH of the solution, we need to find the pKa value of HF, which is the negative logarithm (base 10) of the acid dissociation constant (Ka) for HF. The pKa value for HF is approximately 3.17.
Since the concentration of HF is greater than the concentration of NaOH (0.57 M > 0.13 M), we can consider HF as the predominant acid in the solution.
The Henderson-Hasselbalch equation can be used to calculate the pH of a solution containing a weak acid and its conjugate base:
pH = pKa + log ([A-] / [HA])
Where [A-] is the concentration of the conjugate base (NaF in this case) and [HA] is the concentration of the weak acid (HF).
Since the stoichiometry of the reaction is 1:1, the final concentration of NaF will be equal to the moles of NaOH added.
Final concentration of NaF = Moles of NaOH added / volume of solution
= 0.13 mol / 1.00 L
= 0.13 M
Now, let's substitute the values into the Henderson-Hasselbalch equation:
pH = 3.17 + log (0.13 M / 0.44 M)
= 3.17 + log (0.2955)
= 3.17 + (-0.53)
= 2.64
Therefore, the correct answer to three significant figures is pH = 2.64.
You likely obtained a different answer because you might have used incorrect concentrations or made an error in the calculation.