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Chapter 15: Problem 93

A certain acetic acid solution has \(\mathrm{pH}=2.68\). Calculate the volume of \(0.0975 M \mathrm{KOH}\) required to reach the equivalence point in the titration of \(25.0 \mathrm{~mL}\) of the acetic acid solution.

Short Answer

Expert verified
To find the volume of 0.0975 M KOH required for titration, first, calculate the concentration of acetic acid with a given pH of 2.68: \[[\text{H}^+] = 10^{-2.68}\] Then, find the moles of acetic acid in the 25.0 mL solution: \[\text{moles of CH}_3\text{COOH} = (10^{-2.68})(0.025 \text{L})\] Since one mole of KOH reacts with one mole of CH\(_3\)COOH, the moles of KOH required for equivalence point are equal to the moles of CH\(_3\)COOH. Finally, calculate the volume of KOH needed: \[\text{volume of KOH} = \frac{\text{moles of CH}_3\text{COOH}}{0.0975 M}\]

Step by step solution

01

Calculate the concentration of acetic acid using pH

The formula to find the concentration of acetic acid using pH is: \[\text{pH} = -\log [\text{H}^+]\] We have a pH of 2.68, so we can solve for the concentration of hydrogen ions: \[[\text{H}^+] = 10^{-\text{pH}}\] \[[\text{H}^+] = 10^{-2.68}\]
02

Find the initial concentration of acetic acid

Since acetic acid is a weak acid, it does not ionize completely in water. However, in this case, we'll assume the initial concentration of \(\mathrm{H}^+\) ions is equal to the initial concentration of acetic acid \([\text{CH}_3\text{COOH}]\), as titration with a strong base would force the reaction to completion. Therefore, we have: \[[\text{CH}_3\text{COOH}] = [\text{H}^+]\] \[[\text{CH}_3\text{COOH}] = 10^{-2.68}\]
03

Using stoichiometry to find moles of acetic acid

Now that we have found the concentration of acetic acid in the solution, we can find the moles of acetic acid in the 25.0 mL solution: \[\text{moles of CH}_3\text{COOH} = (\text{concentration})\times(\text{volume})\] \[\text{moles of CH}_3\text{COOH} = (10^{-2.68})(0.025 \text{L})\]
04

Find moles of KOH required for equivalence point

At the equivalence point, one mole of \(\mathrm{KOH}\) will react with one mole of \(\mathrm{CH}_3\text{COOH}\). Therefore, the moles of \(\mathrm{KOH}\) required for the equivalence point will be equal to the moles of \(\mathrm{CH}_3\text{COOH}\): \[\text{moles of KOH} = \text{moles of CH}_3\text{COOH}\]
05

Calculate the volume of KOH needed for titration

We know the concentration of the \(\mathrm{KOH}\) solution (0.0975 M) and the moles of \(\mathrm{KOH}\) required for the equivalence point. We can now find the volume of \(\mathrm{KOH}\) needed: \[\text{volume of KOH} = \frac{\text{moles of KOH}}{\text{concentration of KOH}}\] \[\text{volume of KOH} = \frac{\text{moles of CH}_3\text{COOH}}{0.0975 M}\] Now, use the moles of CH\(_3\)COOH calculated in Step 3 to find the volume of KOH required for the titration.

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Most popular questions from this chapter

You make \(1.00 \mathrm{~L}\) of a buffered solution \((\mathrm{pH}=4.00)\) by mixing acetic acid and sodium acetate. You have \(1.00 M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

a. Calculate the \(\mathrm{pH}\) of a buffered solution that is \(0.100 \mathrm{M}\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 \mathrm{M}\) 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 $$ \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q) $$ to calculate the \(\mathrm{pH}\). c. Do the same as in part \(\mathrm{b}\), but use the following equilibrium to calculate the \(\mathrm{pH}\) : $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q) $$ d. Do your answers in parts \(\mathrm{b}\) and \(\mathrm{c}\) agree? Explain.

Calculate the \(\mathrm{pH}\) of a solution that is \(0.60 \mathrm{M} \mathrm{HF}\) and \(1.00 \mathrm{M} \mathrm{KF}\).

One method for determining the purity of aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is to hydrolyze it with NaOH solution and then to titrate the remaining \(\mathrm{NaOH}\). The reaction of aspirin with \(\mathrm{NaOH}\) is as follows: \(\mathrm{C}_{?} \mathrm{H}_{3} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) Aspirin $$ \begin{array}{c} \text { Bail } \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { Salicylate ion Acetate ion } \end{array} $$ A sample of aspirin with a mass of \(1.427 \mathrm{~g}\) was boiled in \(50.00 \mathrm{~mL}\) of \(0.500 M \mathrm{NaOH}\). After the solution was cooled, it took \(31.92 \mathrm{~mL}\) of \(0.289 \mathrm{M} \mathrm{HCl}\) to titrate the excess \(\mathrm{NaOH}\). Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

Sketch the titration curve for the titration of a generic weak base B with a strong acid. The titration reaction is $$ \mathrm{B}+\mathrm{H}^{+} \rightleftharpoons \mathrm{BH}^{+} $$ On this curve, indicate the points that correspond to the following: a. the stoichiometric (equivalence) point b. the region with maximum buffering c. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\)
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