You have a solution of the weak acid HA and add some HCl to it. What are the major species in the solution? What do you need to know to calculate the 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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After adding HCl to the weak acid HA solution, the major species present are HA, A-, H3O+, Cl-, and H2O. To calculate the pH of the solution, we need the initial concentrations of HA and HCl, and the Ka of HA. We can use the Henderson-Hasselbalch equation \(pH = pKa + log(\frac{[A^-]}{[HA]})\) to calculate the pH. To compare the pH of the solution of just HA with the final mixture, calculate the pH of both solutions and compare the values. The pH of the final mixture is expected to be lower (more acidic) than the pH of the solution of just HA.

Step by step solution

01

Identify the major species

After adding HCl to the weak acid HA solution, we have the following species present in the solution: HA (weak acid), A- (conjugate base resulting from the ionization of HA), H3O+ (hydronium ions), Cl- (chloride ions from HCl), and H2O (water). HCl is a strong acid and it will dissociate completely, so the concentration of Cl- will be equal to the concentration of HCl added. The concentration of H3O+ will be the sum of the H3O+ generated by the dissociation of both HA and HCl.
02

Determine the information needed to calculate the pH

To calculate the pH of the solution, we need to know the initial concentrations of HA and HCl, as well as the acid dissociation constant (Ka) of HA. This information will be necessary to apply the Henderson-Hasselbalch equation, which is given by: \(pH = pKa + log(\frac{[A^-]}{[HA]})\) Where \(pH\) is the pH of the solution, \(pKa\) is the negative logarithm of the acid dissociation constant (Ka), \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the undissociated weak acid.
03

Calculate the pH of the solution

Once we have the initial concentrations of HA and HCl and the Ka of HA, we can calculate the pH of the solution using the following procedure: 1. Calculate the initial concentration of H3O+ by considering the complete dissociation of HCl, which gives \([H3O+] = [HCl]\). 2. Calculate the pKa of HA using the equation \(pKa = -log(Ka)\). 3. Calculate the concentration of A- in the equilibrium using the following relationship: \([A^-] = \frac{[H3O+]}{10^{(pKa - pH)}}\) 4. Calculate the concentration of undissociated HA using the expression: \([HA] = [HA]_0 - [A^-]\) Where \([HA]_0\) is the initial concentration of HA before the addition of HCl. 5. Finally, use the obtained values of [A-] and [HA] in the Henderson-Hasselbalch equation to calculate the pH of the solution.
04

Compare the pH of the solution of just HA with the final mixture

In a solution containing just HA, we can calculate its pH using the Henderson-Hasselbalch equation as well with the initial concentrations of HA and A-. After computing the pH of the final mixture (after the addition of HCl), compare the two pH values to determine which solution is more acidic or alkaline. Since HCl is a strong acid, we can expect that the pH of the final mixture will be lower (more acidic) than the pH of the solution of just HA.

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Most popular questions from this chapter

Derive an equation analogous to the Henderson-Hasselbalch equation but relating pOH and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\)

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 0.125\(M \mathrm{NaOH}\) , requiring 16.00 \(\mathrm{mL}\) to reach the equivalence point. During the titration, the pH after adding 2.00 \(\mathrm{mL}\) . NaOH was 6.912 . Calculate \(K_{\mathrm{a}}\) for the weak acid.

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 \(\mathrm{L}\) of each of the following solutions: a. 0.050\(M\) propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)\( and 0.080\)M$ sodium propanoate b. 0.50\(M\) propanoic acid and 0.80\(M\) sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Which of the following can be classified as buffer solutions? $$ \begin{array}{l}{\text { a. } 0.25 M \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}} \\ {\text { b. } 0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}} \\ {\text { c. } 0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{MKOCl}}\end{array} $$ $$ \begin{array}{l}{\text { d. } 0.70 M \mathrm{KOH}+0.70 \mathrm{M} \text { HONH_ }} \\ {\text { e. } 0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}}\end{array} $$

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intra- cellular fluids at pH values generally between 7.1 and \(7.2 .\) a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) inintracellular fluid at \(\mathrm{pH}=7.15 ?\) $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$ b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the pH of intracellular fluid? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$

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