You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the \(\mathrm{pH}\) of the solution, and how would you use this information? How does the \(\mathrm{pH}\) of the solution of just the HA compare with that of the final mixture? Explain.

When a weak acid HA is mixed with the salt NaA, the major species in the solution are: 1. HA (the weak acid) 2. A- (the conjugate base of the weak acid, coming from the salt NaA) 3. H2O (water, as the solvent) The ions Na+ from the salt, do not affect the pH, so they are not considered in this analysis.

To calculate the pH of this solution, we need to determine the equilibrium concentrations of the major species (HA, A-, and H3O+). For this, we need the following information: 1. Initial concentration of HA and NaA (which gives the initial A- concentration) 2. The acid dissociation constant, Ka, of the weak acid HA

Once we have the initial concentrations and the Ka, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution: \[pH = pK_a + log \frac{[A^-]}{[HA]}\] where: - pH is the pH of the solution - pKa = -log(Ka) is the negative logarithm of the acid dissociation constant - [A-] is the equilibrium concentration of the conjugate base - [HA] is the equilibrium concentration of the weak acid Using the Ka value and the initial concentrations, we can solve the equation to find the pH of the solution.

The pH of the solution of just the weak acid (HA) is given by the pH value that we would have if no salt (NaA) was added: \[pH_{HA} = \frac{-log(K_a)}{2}\] When some of the salt NaA is added to it, the pH of the final mixture becomes: \[pH_{mixture} = pK_a + log \frac{[A^-]}{[HA]}\] Comparing the two expressions, we observe that the second term of the Henderson-Hasselbalch equation, \(log \frac{[A^-]}{[HA]}\), reflects the effect of added salt on pH. Since the concentration of the conjugate base A- increases with the addition of NaA, and HA remains the same, the term \(\frac{[A^-]}{[HA]}\) is greater than 1, resulting in a positive value for the logarithm. Therefore, the pH of the final mixture is higher than the pH of the weak acid HA alone: \[pH_{mixture} > pH_{HA}\] This indicates that the addition of the salt NaA increases the pH of the solution, making it less acidic.