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Chapter 15: Problem 108

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NaOH}\) . The student then adds 13.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) . The pH of the resulting solution is \(4.70 .\) How is the value of \(\mathrm{p} K_{2}\) for the unknown acid related to 4.70\(?\)

Short Answer

Expert verified
The value of \(pK_2\) for the unknown weak acid, HA, is related to the pH value of 4.70 through the Henderson-Hasselbalch equation: \(pK_2 = 4.70 - log(\frac{0.0316}{0.100})\).

Step by step solution

01

Determine the moles of NaOH used in the titration

To calculate the moles of NaOH, we will use the formula: Moles of NaOH = volume of NaOH (L) × concentration of NaOH (M) The given volume and concentration of NaOH are 25.0 mL and 0.100 M, respectively. First, we have to convert the volume from mL to L, so: 25.0 mL = 0.025 L Now, we can calculate the moles of NaOH: Moles of NaOH = 0.025 L × 0.100 M = 0.00250 mol
02

Determine the moles of HA titrated

Because the reaction between the weak acid (HA) and the NaOH is 1:1, the moles of titrated HA will be equal to the moles of NaOH: Moles of HA = 0.00250 mol
03

Determine the moles of added HCl and the moles of NaOH that remain

We need to calculate the moles of added HCl using the same formula as before: Moles of HCl = volume of HCl (L) × concentration of HCl (M) The given volume and concentration of HCl are 13.0 mL and 0.100 M, respectively. We have to convert the volume from mL to L: 13.0 mL = 0.013 L Now, we can calculate the moles of HCl: Moles of HCl = 0.013 L × 0.100 M = 0.00130 mol
04

Calculate the moles of resulting HA and A⁻ (conjugate base)

Since HCl is a strong acid, it will react with the conjugate base (A⁻) produced during the titration: A⁻ + HCl → HA The remaining moles of A⁻ after the addition of HCl: Moles of A⁻ = moles of NaOH - moles of HCl = 0.00250 mol - 0.00130 mol = 0.00120 mol The moles of HA will increase by the amount of HCl added: Moles of HA = 0.00250 mol + 0.00130 mol = 0.00380 mol
05

Determine the concentrations of HA and A⁻

The total volume of the solution after the addition of HCl is: Total volume = 25.0 mL + 13.0 mL = 38.0 mL = 0.038 L Now we can calculate the concentrations: Concentration of HA = moles of HA / total volume = 0.00380 mol / 0.038 L = 0.100 M Concentration of A⁻ = moles of A⁻ / total volume = 0.00120 mol / 0.038 L = 0.0316 M
06

Use the Henderson-Hasselbalch equation to find the relation between pK₂ and pH

The Henderson-Hasselbalch equation is: \(pH = pK_{2} + log(\frac{[A^-]}{[HA]})\) We know that the pH of the resulting solution is 4.70: 4.70 = pK₂ + log(\(\frac{0.0316}{0.100}\)) To find the relation between pK₂ and the given pH value, we simply need to solve the equation for pK₂: pK₂ = 4.70 - log(\(\frac{0.0316}{0.100}\)) Therefore, the value of pK₂ for the unknown acid is related to the pH value 4.70 through the above equation.

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Most popular questions from this chapter

Which of the following can be classified as buffer solutions? $$ \begin{array}{l}{\text { a. } 0.25 M \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}} \\ {\text { b. } 0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}} \\ {\text { c. } 0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{MKOCl}}\end{array} $$ $$ \begin{array}{l}{\text { d. } 0.70 M \mathrm{KOH}+0.70 \mathrm{M} \text { HONH_ }} \\ {\text { e. } 0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}}\end{array} $$

Calculate the volume of \(1.50 \times 10^{-2} M \mathrm{NaOH}\) that must be added to 500.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15 .\)

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.

Consider a buffered solution containing $\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\( and \)\mathrm{CH}_{3} \mathrm{NH}_{2}$ . Which of the following statements concerning this solution is(are) true? $\left(K_{\mathrm{a}} \text { for } \mathrm{CH}_{3} \mathrm{NH}_{3}+=2.3 \times 10^{-11}\right)$ a. A solution consisting of 0.10$M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\( and 0.10 \)\mathrm{M}\( \)\mathrm{CH}_{3} \mathrm{NH}_{2}$ would have a greater buffering capacity than one containing 1.0 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and 1.0 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) b. If $\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right],\( then the \)\mathrm{pH}$ is larger than the \(\mathrm{p} K_{\mathrm{a}}\) value. c. Adding more \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\right]\) to the initial buffer solution will decrease the pH. d. If $\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]<\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right],\( then \)\mathrm{pH}<3.36$ e. If $\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right],\( then \)\mathrm{pH}=10.64$

Derive an equation analogous to the Henderson-Hasselbalch equation but relating pOH and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\)
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