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Chapter 15: Problem 92

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?

Short Answer

Expert verified
To make 1.00 L of a buffered solution with a pH of 4.00, mix approximately 0.854 L (854 mL) of a 1.00 M acetic acid solution and 0.146 L (146 mL) of a 1.00 M sodium acetate solution.

Step by step solution

01

Write down the Henderson-Hasselbalch equation

The equation is given by: \(pH = pKa + log10(\frac{[A-]}{[HA]})\) Here, pH is the desired pH of the buffered solution, pKa is the acidity constant of the weak acid, [HA] is the molar concentration of the weak acid, and [A-] is the molar concentration of the conjugate base.
02

Find the pKa of acetic acid

The pKa value of acetic acid is 4.76. We will use this value in the Henderson-Hasselbalch equation to find the ratio of [A-]/[HA].
03

Plug in the given pH value and pKa value into the equation

Given, pH = 4.00 and pKa = 4.76. Substituting these values into the equation, we get: \(4.00 = 4.76 + log10(\frac{[A-]}{[HA]})\)
04

Solve for the ratio of [A-]/[HA]

Subtract 4.76 from both sides: \(-0.76 = log10(\frac{[A-]}{[HA]})\) Now, take the antilogarithm of both sides: \(\frac{[A-]}{[HA]} = 10^{-0.76}\) Evaluate the expression on the right: \(\frac{[A-]}{[HA]} = 0.17\)
05

Calculate the volume of each solution

We know that the total volume of the buffered solution is 1.00 L. Let's assume that we mix V1 liters of acetic acid (1.00 M) and V2 liters of sodium acetate (1.00 M) to get this volume. Then: \(V1 + V2 = 1.00 \, L\) Using the [A-]/[HA] ratio, we can write the following equation: \(\frac{1.00*V2}{1.00*V1} = 0.17\) Now, we have two equations and two variables (V1 and V2).
06

Solve for V1 and V2

From the second equation, we get \(V2 = 0.17*V1\) Substituting this expression in the first equation \(V1 + 0.17*V1 = 1.00 \, L\) \(1.17*V1 = 1.00 \, L\) Divide both sides by 1.17 to get the value of V1: \(V1 = \frac{1.00}{1.17} \, L \approx 0.854 \, L\) Now, substitute the value of V1 back into the second equation to get the value of V2: \(V2 = 0.17*0.854 \, L \approx 0.146 \, L\)
07

Provide the answer

To make 1.00 L of a buffered solution with a pH of 4.00, we should mix approximately 0.854 L (854 mL) of a 1.00 M acetic acid solution and 0.146 L (146 mL) of a 1.00 M sodium acetate solution.

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Most popular questions from this chapter

Carbonate buffers are important in regulating the pH of blood at \(7.40 .\) If the carbonic acid concentration in a sample of blood is 0.0012 M, determine the bicarbonate ion concentration required to buffer the pH of blood at pH \(=7.40\) $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7} $$

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 \(\mathrm{L}\) of each of the following solutions: a. 0.050\(M\) propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)\( and 0.080\)M$ sodium propanoate b. 0.50\(M\) propanoic acid and 0.80\(M\) sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

In the titration of 50.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) methylamine, CH_{3} \mathrm { NH } _ { 2 } \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with 0.50 $\mathrm{M} \mathrm{HCl}$ , calculate the pH under the following conditions. a. after 50.0 \(\mathrm{mL}\) of 0.50\(M \mathrm{HCl}\) has been added b. at the stoichiometric point

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