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Chapter 17: Problem 57

The pH of a solution of \(19.5 \mathrm{g}\) of malonic acid in \(0.250 \mathrm{L}\) is \(1.47 .\) The pH of a \(0.300 \mathrm{M}\) solution of sodium hydrogen malonate is 4.26. What are the values of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for malonic acid?

Short Answer

Expert verified
In order to find the values of \(K_{a_{1}}\) and \(K_{a_{2}}\), first, calculate the concentration of H+ ions from the given pH values. Then, calculate the molarity of the malonic acid using the weight and volume of the solution. Finally, use these values to calculate \(K_{a_{1}}\) and \(K_{a_{2}}\) using their respective formulas. The values will depend upon the specific numerical values obtained during the calculations.

Step by step solution

01

Identifying the Given Information

Identify all the given data and write them down. The weight of the malonic acid is given to be 19.5g, volume of the solution: 0.250L, pH of the solution: 1.47. Furthermore, pH of 0.300M solution of sodium hydrogen malonate is 4.26.
02

Using the Formula for pH

For each solution, find the concentration of H+ ions (\([H^+]\)) using the formula \(pH = -log [H^+]\). For solution 1, \( [H^+] = 10^{-pH} = 10^{-1.47} \). For solution 2, \([H^+] = 10^{-pH} = 10^{-4.26}\)
03

Determine the Molar Mass of Malonic Acid

Using the periodic table, find the molar mass of malonic acid(C3H4O4) to be approximately 104g/mol.
04

Find the Concentration of the Malonic Acid

Using the formula \(Molarity = \frac{Weight}{Molar Mass \times Volume}\), find the concentration of the malonic acid. The molar mass is derived from Step 3, molarity of malonic acid M1 = \( \frac{19.5g}{104g/mol \times 0.250L} \)
05

Determine the Acid Ionization Constants \(K_{a_{1}}\)

Use the formula for finding the acid ionization constant \(K_{a_{1}}\). \( K_{a_{1}} = \frac{[H^+][Hmal^-]}{[Hmal]} = \frac{[H^+]^2}{M1 - [H^+]}\) Where: [Hmal^-] and [H^+] are approximately equal as for every mole of malonic acid that ionizes, one mole of H ions and one mole of Hmal- ions are produced. Thus their concentrations are equal. Furthermore, [Hmal] = M1 - [H^+] because the initial concentration of the malonic acid is M1 and as it ionizes, its concentration decreases by the same amount as [H^+].
06

Determine the Acid Ionization Constants \(K_{a_{2}}\)

From Step 2, the [H+] and hence [Mal^2-] for the second ionization may be calculated for the sodium hydrogen malonate solution. Therefore, using the equation \( K_{a_{2}} = \frac{[H^+][Mal^{2-}]}{[Hmal^-]} = \frac{[H^+]^2}{M2 - [H^+]}\). Therefore, we can calculate \(K_{a_{2}}\) by substituting the known values.

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Most popular questions from this chapter

Two aqueous solutions are mixed: \(50.0 \mathrm{mL}\) of 0.0150 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(50.0 \mathrm{mL}\) of \(0.0385 \mathrm{M} \mathrm{NaOH} .\) What is the pH of the resulting solution?

A handbook lists various procedures for preparing buffer solutions. To obtain a \(\mathrm{pH}=9.00,\) the handbook says to mix \(36.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NH}_{3}\) with \(64.00 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (a) Show by calculation that the pH of this solution is 9.00. (b) Would you expect the \(\mathrm{pH}\) of this solution to remain at \(\mathrm{pH}=9.00\) if the \(100.00 \mathrm{mL}\) of buffer solution were diluted to 1.00 L? To 1000 L? Explain. (c) What will be the pH of the original \(100.00 \mathrm{mL}\) of buffer solution if \(0.20 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) is added to it? (d) What is the maximum volume of \(1.00 \mathrm{M} \mathrm{HCl}\) that can be added to \(100.00 \mathrm{mL}\) of the original buffer solution so that the pH does not drop below \(8.90 ?\)

In the titration of \(20.00 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH},\) calculate the number of milliliters of \(0.200 \mathrm{M} \mathrm{HCl}\) that must be added to reach a pH of (a) \(12.55,\) (b) \(10.80,\) (c) 4.25

Even though the carbonic acid-hydrogen carbonate buffer system is crucial to the maintenance of the \(\mathrm{pH}\) of blood, it has no practical use as a laboratory buffer solution. Can you think of a reason(s) for this?

You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all \(0.100 \mathrm{M},\) are available to you: HCOOH, CH \(_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{NaCHOO}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{4} . \quad\) Describe how you would prepare this buffer solution. [Hint: What volumes of which solutions would you use?
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