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Chapter 15: Problem 45

Some \(\mathrm{K}_{2} \mathrm{SO}_{3}\) and \(\mathrm{KHSO}_{3}\) are dissolved in 250.0 \(\mathrm{mL}\) of solution and the resulting \(\mathrm{pH}\) is \(7.25 .\) Which is greater in this buffer solution, the concentration of \(\mathrm{SO}_{3}^{2-}\) or the concentration of \(\mathrm{HSO}_{3}-7\) If \(\left[\mathrm{SO}_{3}^{2-}\right]=1.0 \mathrm{M}\) in this solution, calculate the concentration of \(\mathrm{HSO}_{3}\) .

Short Answer

Expert verified
In conclusion, the concentration of SO3^2- is greater than the concentration of HSO3- in the given buffer solution, and the concentration of HSO3- is 0.316 M.

Step by step solution

01

Write down the Henderson-Hasselbalch equation for a buffer solution

The Henderson-Hasselbalch equation for a buffer solution is given by: pH = pKa + log\(\frac{[\text{A}^-]}{[\text{HA}]}\) Where pH is the solution's pH, pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid, and [\text{A}^-] and [\text{HA}] are the molar concentrations of the conjugate acid/base pair. In our case, the weak acid is HSO3- and its conjugate base is SO3^2-.
02

Calculate pKa using the given pH value

We are given that the pH of the solution is 7.25. To find the pKa of HSO3-, we can use the relationship between pH, pOH, pKw, pKa, and pKb: pOH = 14 - pH = 14 - 7.25 = 6.75 We can say \(pK_a = p_{OH}\), That gives, \(pK_a = 6.75\)
03

Use the Henderson-Hasselbalch equation to find the concentration of HSO3-

Now we can plug the values of pH, pKa, and [\(\text{SO}_{3}^{2-}\)] into the Henderson-Hasselbalch equation: 7.25 = 6.75 + log\(\frac{[\text{SO}_{3}^{2-}]}{[\text{HSO}_{3}^{-}]}\) Rearrange the equation to solve for [\(\text{HSO}_{3}^{-}\)]: [\(\text{HSO}_{3}^{-}\)] = [\(\text{SO}_{3}^{2-}\)] / 10^(\(pH - pK_a\)) [\(\text{HSO}_{3}^{-}\)] = \(1.0 M\) / 10^(\(7.25 - 6.75\)) [\(\text{HSO}_{3}^{-}\)] = \(1.0 M\) / 10^(\(0.5\)) [\(\text{HSO}_{3}^{-}\)] = 0.316 M
04

Compare the concentrations and determine which ion has a higher concentration

Now that we have the concentration of HSO3- (0.316 M), we can compare it to the given concentration of SO3^2- (1.0 M): [\(\text{SO}_{3}^{2-}\)] = 1.0 M [\(\text{HSO}_{3}^{-}\)] = 0.316 M Since the concentration of SO3^2- (1.0 M) is greater than the concentration of HSO3- (0.316 M), the concentration of SO3^2- is greater in this buffer solution. In conclusion, the concentration of SO3^2- is greater than the concentration of HSO3- in the given buffer solution, and the concentration of HSO3- is 0.316 M.

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Most popular questions from this chapter

A \(10.00-\mathrm{g}\) sample of the ionic compound \(\mathrm{NaA},\) where \(\mathrm{A}^{-}\) is the anion of a weak acid, was dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution and was then titrated with $0.100 \mathrm{M}\( HCl. After \)500.0 \mathrm{~mL}\( HCl was added, the \)\mathrm{pH}$ was 5.00 . The experimenter found that \(1.00 \mathrm{~L}\) of $0.100 \mathrm{M} \mathrm{HCl}$ was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.

Consider a solution formed by mixing 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}, 30.0 \mathrm{mL}\) of 0.100 \(\mathrm{M}\) HOCl, 25.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) \(\mathrm{NaOH}, 25.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2},\) and 10.0 \(\mathrm{mL}\) of 0.150 \(M \mathrm{KOH}\) . Calculate the \(\mathrm{pH}\) of this solution.

Calculate the \(\mathrm{pH}\) of a solution that is 1.00 \(\mathrm{M}\) HNO, and 1.00 \(\mathrm{M} \mathrm{NaNO}_{2}\)

Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to 1.0 \(\mathrm{L}\) of 1.0 $\mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ to produce a solution buffered at each pH. $$ \text{(a)}\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} \quad \text { b. } \mathrm{pH}=4.20 \quad \text { c. } \mathrm{pH}=5.00 $$

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100\(M\) lactic actid $\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right.\( \)\mathrm{p} K_{\mathrm{a}}=3.86 )$ is titrated with 0.100 \(\mathrm{M}\) NaOH solution. Calculate the pH after the addition of $0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL},\( \)20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\( \)26.0 \mathrm{mL}, 28.0 \mathrm{mL}$ , and 30.0 \(\mathrm{mL}\) of the NaOH. Plot the results of your calculations as \(\mathrm{pH}\) versus milliliters of NaOH added.
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