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Chapter 15: Problem 118

Malonic acid \(\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\) is a diprotic acid. In the titration of malonic acid with \(\mathrm{NaOH}\), stoichiometric points occur at \(\mathrm{pH}=3.9\) and \(8.8\). A \(25.00-\mathrm{mL}\) sample of malonic acid of unknown concentration is titrated with \(0.0984 \mathrm{M} \mathrm{NaOH}\), requiring \(31.50 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (See Exercise 113.)

Short Answer

Expert verified
The concentration of the initial malonic acid solution is approximately \(0.0621 \, M\).

Step by step solution

01

Analyze the titration reaction

Let's first write down the balanced chemical equation for the titration reaction happening at the second stoichiometric point (pH=8.8): \(HO_2CCH_2CO_2H + 2 NaOH\longrightarrow Na_2[H_2O_2CCH_2CO_2]\) At the second stoichiometric point, one molecule of diprotic malonic acid reacts with two molecules of NaOH.
02

Calculate the moles of NaOH used in the reaction

We know the concentration and volume of NaOH solution used for titration. Using this information, we can calculate the moles of NaOH. Moles of NaOH = Molarity × Volume The volume must be converted to liters: Volume of NaOH = \(31.50 \, mL = 31.50 \times 10^{-3} \, L\) Moles of NaOH = \(0.0984\,M \times 31.50 \times 10^{-3} \, L = 3.10464 \times 10^{-3} \, moles\)
03

Calculate the moles of malonic acid

In the balanced chemical equation, we see that the one molecule of malonic acid reacts with two molecules of NaOH. Hence, we can simply divide the moles of NaOH by 2 to obtain the moles of malonic acid: Moles of malonic acid = Moles of NaOH / 2 Moles of malonic acid = \(3.10464 \times 10^{-3} \, moles / 2 = 1.55232 \times 10^{-3} \, moles\)
04

Calculate the concentration of the initial malonic acid solution

Finally, we can find the concentration of the initial malonic acid solution by dividing the moles of malonic acid by its volume: Concentration of Malonic Acid = Moles of Malonic Acid / Volume of Malonic Acid Solution The volume must be converted to liters: Volume of Malonic Acid Solution = \(25.00\, mL = 25.00 \times 10^{-3} L\) Concentration of Malonic Acid = \(1.55232 \times 10^{-3} \, moles / 25.00 \times 10^{-3} \, L = 0.0620928 \, M\) The concentration of the initial malonic acid solution is approximately \(0.0621 \, M\).

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Most popular questions from this chapter

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.70\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to 4.70?

Consider \(1.0 \mathrm{~L}\) of a solution that is \(0.85 \mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\) and \(0.80 \mathrm{M} \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} \cdot\right)\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after \(0.10 \mathrm{~mole}\) of \(\mathrm{HCl}\) has been added to the original solution. Assume no volume change on addition of \(\mathrm{HCl}\). c. Calculate the \(\mathrm{pH}\) after \(0.20 \mathrm{~mole}\) of \(\mathrm{NaOH}\) has been added to the original buffer solution. Assume no volume change on addition of \(\mathrm{NaOH}\).

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like \(\mathrm{NaOH}\) is added, the HA reacts with the \(\mathrm{OH}^{-}\) to form \(\mathrm{A}^{-}\). Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding \(\mathrm{HCl}\) to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right) .\) Thus how can we claim that a buffered solution resists changes in the \(\mathrm{pH}\) of the solution?" How would you explain buffering to this friend?

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After \(23.75 \mathrm{~mL}\) of the weak acid solution has been added to \(50.0 \mathrm{~mL}\) of the \(0.100 \mathrm{M} \mathrm{NaOH}\) solution, the \(\mathrm{pH}\) of the resulting solution is \(10.50 .\) Calculate the original concentration of the solution of weak acid.

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?
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