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Chapter 14: Problem 84

You make 1.00 L of a buffered solution \((p H=4.00)\) 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?

Short Answer

Expert verified
To make a 1.00 L buffered solution with a pH of 4.00, mix approximately \(0.155 \,L\) of sodium acetate solution and \(0.85 \,L\) of acetic acid solution.

Step by step solution

01

Write down the Henderson-Hasselbalch equation for the system

The Henderson-Hasselbalch equation is given by: \( pH = pK_a + log \frac{[A^-]}{[HA]} \) Where, - pH is the required pH of the buffered solution, - pK_a is the acid dissociation constant corresponding to the acid, - [A^-] is the concentration of the conjugate base (acetate ion, CH3COO-), - [HA] is the concentration of the weak acid (acetic acid, CH3COOH).
02

Find the pK_a value for acetic acid

The pK_a value for acetic acid is 4.74. We will use this value in the Henderson-Hasselbalch equation.
03

Substitute pH and pKa values into the equation

Since we need a pH of 4.00, we can plug the values into the equation: \( 4.00 = 4.74 + log\frac{[CH3COO^-]}{[CH3COOH]} \)
04

Rearrange the equation to find the ratio

We need to rearrange the equation to find the ratio between concentrations of acetic acid and sodium acetate: \( log\frac{[CH3COO^-]}{[CH3COOH]} = 4.00 - 4.74 \) \( log\frac{[CH3COO^-]}{[CH3COOH]} = -0.74 \) Taking the antilog to remove the log: \( \frac{[CH3COO^-]}{[CH3COOH]} = 10^{-0.74} \) \( \frac{[CH3COO^-]}{[CH3COOH]} ≈ 0.183 \)
05

Use the relationship between the volumes and the ratio

Let the volume of the acetate solution be V_a and the volume of acetic acid solution be V_ac. Since the solution is 1.00 L and their concentrations are equal (1.00 M), we have: \( \frac{V_a}{V_ac} = 0.183 \)
06

Use the total volume to find the individual volumes

We know that the total volume (V_total) of the buffered solution is 1.00 L: \( V_a + V_ac = V_{total} = 1.00 \,L \) Now we can substitute the ratio from Step 5: \( V_a = 0.183 V_ac \) And solve for the volumes: \( 0.183 V_ac + V_ac = 1.00 \,L \) \( V_ac(0.183 + 1) = 1.00 \,L \) \( V_ac ≈ 0.85 \,L \) Now using the ratio between the volumes, we find the volume of acetate solution: \( V_a = 0.183 V_ac = 0.183 × 0.85 \,L \) \( V_a ≈ 0.155 \,L \)
07

Final Answer

To make a 1.00 L buffered solution with a pH of 4.00, we need to mix approximately 0.155 L of sodium acetate solution and 0.85 L of acetic acid solution.

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Most popular questions from this chapter

Calculate the pH of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 M\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

What concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is necessary to buffer a \(0.52-M\) \(\mathrm{NH}_{3}\) solution at \(\mathrm{pH}=9.00 ?\left(K_{\mathrm{b}} \text { for } \mathrm{NH}_{3}=1.8 \times 10^{-5} .\right)\)

Consider the titration of \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HI by \(0.250 \mathrm{M}\) NaOH. a. Calculate the \(\mathrm{pH}\) after \(20.0 \mathrm{mL}\) of \(\mathrm{NaOH}\) has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=\) \(7.00 ?\)

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

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