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 pH of the solution of just the HA compare with that of the final mixture? Explain.

Short Answer

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The major species in the solution are HA, A-, Na+, and H+. To calculate the pH, we need the initial concentrations of HA and A- and the value of Ka for the weak acid HA. We can then use the equilibrium expression \(K_a = \frac{[H^+]^2}{[HA]-[H^+]}\) to solve for [H+] and calculate the pH. The pH of the initial solution (just HA) will be higher than that of the final mixture (HA and NaA) due to the buffering effect of the added conjugate base A-.

Step by step solution

01

Identify major species in the solution

Since we start with a solution of weak acid HA and add salt NaA, the major species in the initial solution are the weak acid HA, its conjugate base A-, and the cations Na+ and H+. When the salt NaA is added, it will dissociate completely into Na+ and A- ions because it is a strong electrolyte. The major species in the final solution are therefore HA, A-, Na+, and H+.
02

Determine information needed to calculate pH

To calculate the pH of the solution, we need to determine the concentration of H+ ions. This can be obtained by analyzing the equilibrium reaction between the weak acid HA and its conjugate base A-, using the acid dissociation constant (Ka) for the weak acid and the concentrations of the major species. To perform this calculation, we need to know the initial concentrations of HA and A- (after adding NaA) and the value of Ka for the weak acid HA.
03

Calculate the pH of the solution using the given information

Using the initial concentrations of HA and A- and the Ka value for the weak acid, we can set up the following equilibrium expression: \[K_a = \frac{[H^+][A^-]}{[HA]}\] Since the concentration of H+ ions and A- ions will be equal at equilibrium, we can rewrite the equilibrium expression as: \[K_a = \frac{[H^+]^2}{[HA]-[H^+]}\] Solving this equation for [H+] will allow us to calculate the pH of the solution using the relationship pH = -log10 [H+].
04

Compare the pH of the initial HA solution and the final mixture

The pH of the initial HA solution will be higher than that of the final mixture, due to the presence of the conjugate base A- from the added salt NaA. This is because A- ions act as a buffer, reducing the change in pH when more acid is added to the solution. As a result, the pH of the solution will be maintained within a narrow range, decreasing the overall acidity of the final mixture compared to the initial HA solution. In summary, the major species in the solution are HA, A-, Na+, and H+. To calculate the pH of the solution, we need to know the initial concentrations of HA and A- and the value of Ka for the weak acid HA. The pH of the initial solution (just HA) will be higher than that of the final mixture (HA and NaA) due to the buffering effect of the added conjugate base A-.

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Most popular questions from this chapter

Calculate the volume of \(1.50 \times 10^{-2} \mathrm{M}\) NaOH that must be added to \(500.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HCl to give a solution that has \(\mathrm{pH}=2.15\).

Two drops of indicator HIn \(\left(K_{\mathrm{a}}=1.0 \times 10^{-9}\right),\) where HIn is yellow and \(\operatorname{In}^{-}\) is blue, are placed in \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HCl. a. What color is the solution initially? b. The solution is titrated with 0.10 \(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after \(200.0 \mathrm{mL}\) NaOH has been added?

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\).

a. Calculate the pH of a buffered solution that is 0.100 \(M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q)$$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the pH: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q)\) d. Do your answers in parts \(b\) and c agree? Explain.

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

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