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.

Short Answer

Expert verified
The given results can be explained as follows: Acid 1 requires 25 mL of 0.200 M NaOH to reach its stoichiometric point, and Acid 2 requires 50 mL of 0.100 M NaOH. Acid 1 has a stronger affinity for NaOH due to its lower pKa1 value (2.98), and its second ionization is negligible due to the high pKa2 value (13.40). For Acid 2, with pKa1 (4.41) and pKa2 (5.28), the second ionization is close enough to influence the reaction but not enough for an additional stoichiometric point to be observed.

Step by step solution

01

Identify given information

Here are the given variables and information in this problem: Acid 1: - pKa1 = 2.98 - pKa2 = 13.40 Acid 2: - pKa1 = 4.41 - pKa2 = 5.28 The "25.00 mL added NaOH" stoichiometric point and the "50.00 mL added NaOH" stoichiometric point both correspond to 5.00 mmol of acid. The concentration of NaOH is 0.200 M.
02

Calculate the moles of each acid

Since we have 5.00 mmol of each acid, we can convert this to moles: 5.00 mmol = 0.005 mol
03

Calculate the volume of NaOH needed for each acid's stoichiometric point

To find the volume of NaOH required for each acid to reach its stoichiometric point, we will use the following formula: Volume of NaOH (L) = moles of Acid / concentration of NaOH Acid 1: Volume of NaOH (L) = 0.005 mol / 0.200 M = 0.025 L or 25 mL Acid 2: Volume of NaOH (L) = 0.005 mol / 0.100 M = 0.050 L or 50 mL
04

Match the stoichiometric points to the acids

Now that we have calculated the volume of NaOH required for each acid's stoichiometric point, we can match them to the given stoichiometric points: - Acid 1 reaches its stoichiometric point when 25.00 mL of NaOH is added. - Acid 2 reaches its stoichiometric point when 50.00 mL of NaOH is added. These results show that Acid 1 has a stronger affinity for the NaOH and neutralizes more quickly due to its lower pKa1 value (pKa1=2.98). Conversely, Acid 2 takes longer to fully neutralize due to its higher pKa1 value (pKa1=4.41). Also, Acid 1's second ionization doesn't affect the stoichiometry, as the pKa2 is too high (pKa2=13.40), whereas Acid 2's pKa2 (pKa2=5.28) is close enough to the first pKa to influence the reaction, but not enough for an additional stoichiometric point to be observed.

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Most popular questions from this chapter

Another way to treat data from a pH titration is to graph the absolute value of the change in \(\mathrm{pH}\) per change in milliliters added versus milliliters added (\DeltapH/ \(\Delta \mathrm{mL}\) versus \(\mathrm{mL}\) added). Make this graph using your results from Exercise \(67 .\) What advantage might this method have over the traditional method for treating titration data?

Sketch the titration curve for the titration of a generic weak base \(\mathrm{B}\) 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. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) d. \(\mathrm{pH}\) depends only on \([\mathrm{B}]\) e. \(\mathrm{pH}\) depends only on \(\left[\mathrm{BH}^{+}\right]\) f. \(\mathrm{pH}\) depends only on the amount of excess strong acid added

Consider the titration of 80.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\( by 0.400 \)\mathrm{M} \mathrm{HCl}$ . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of HCl have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}} \\ {\text { c. } 30.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}}\end{array} $$

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A student dissolves 0.0100 mol of an unknown weak base in 100.0 \(\mathrm{mL}\) water and titrates the solution with 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) After 40.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

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