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Chapter 15: Problem 130

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 NaOH, stoichiometric points occur at \(\mathrm{pH}=3.9\) and $8.8 . \mathrm{A} 25.00-\mathrm{mL}$ sample of malonic acid of unknown concentration is titrated with 0.0984 \(\mathrm{M}\) NaOH, requiring 31.50 \(\mathrm{mL}\) of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution.

Short Answer

Expert verified
The concentration of the initial malonic acid solution is approximately 0.12374 M.

Step by step solution

01

Write the balanced chemical equation for the second ionization reaction of malonic acid

At the second stoichiometric point, both protons of malonic acid are neutralized by the \(\mathrm{NaOH}\). The balanced chemical equation for the second ionization reaction is given by: \(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2}^{2-} + \mathrm{H}_{2} \mathrm{O}\)
02

Find moles of NaOH

Next, we need to find the moles of \(\mathrm{NaOH}\) used in the reaction. Use the formula: moles = concentration × volume; where volume should be in liters. Moles of NaOH = (0.0984 mol/L) × (31.50 mL × (1 L / 1000 mL)) Moles of NaOH = 0.0030936 mol
03

Find moles of malonic acid

Since the stoichiometry between the \(\mathrm{NaOH}\) and malonic acid is 1:1 in the second ionization reaction, the moles of malonic acid required are equal to the moles of \(\mathrm{NaOH}\) used. Moles of malonic acid = 0.0030936 mol
04

Calculate the concentration of malonic acid

Now, we'll use the initial volume of the malonic acid solution and the moles of malonic acid to find its concentration. Use the formula: concentration = moles / volume; where volume should be in liters. Concentration of malonic acid = (0.0030936 mol) / (25.00 mL × (1 L / 1000 mL)) Concentration of malonic acid = 0.12374 M Hence, the concentration of the initial malonic acid solution is approximately 0.12374 M.

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Most popular questions from this chapter

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? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$

A certain indicator HIn has a \(\mathrm{p} K_{\mathrm{a}}\) of 3.00 and a color change becomes visible when 7.00\(\%\) of the indicator has been converted to \(\mathrm{In}^{-}\) . At what pH is this color change visible?

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{HCN}$ by 0.100 \(\mathrm{M} \mathrm{KOH}\) at $25^{\circ} \mathrm{C} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN}=6.2 \times 10^{-10} .\right)$ a. Calculate the pH after 0.0 \(\mathrm{mL}\) of KOH has been added. b. Calculate the pH after 50.0 \(\mathrm{mL}\) of KOH has been added. c. Calculate the pH after 75.0 \(\mathrm{mL}\) of KOH has been added. d. Calculate the pH at the equivalence point. e. Calculate the pH after 125 \(\mathrm{mL}\) of KOH has been added.

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{1}}=5.28 $$ In two separate experiments the pH was measured during the titration of 5.00 \(\mathrm{mmol}\) of each acid with 0.200 \(\mathrm{M} \mathrm{NaOH}\) . Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at 25.00 \(\mathrm{mL}\) added NaOH, and in the other experiment the stoichiometric point was at 50.00 \(\mathrm{mL}\) NaOH. Explain these results.

Calculate the pH of a solution prepared by mixing \(250 . \mathrm{mL}\) of 0.174 \(\mathrm{m}\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{g} / \mathrm{mL} )\) with 38.7 \(\mathrm{g}\) of an aqueous solution that is 1.50\(\% \mathrm{NaOH}\) by mass (density \(=\) 1.02 $\mathrm{g} / \mathrm{mL} ) .\left(K_{\mathrm{a}} \text { for } \mathrm{HF}=7.2 \times 10^{-4} .\right)$
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