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Chapter 17: Problem 95

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid $\mathrm{H}_{2} \mathrm{~A}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)$ or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a 1.0 \(M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) 6.50? (Ignore any volume change.)

Short Answer

Expert verified
To obtain a pH 6.50 buffer, add 0.984 L of the 1.0 M NaOH solution to 1.0 L of the 1.0 M H2A solution.

Step by step solution

01

Write the dissociation reactions

The dissociation reactions for the given weak diprotic acid H2A are: \(1st \ dissociation: \mathrm{H}_{2}\mathrm{A\longleftrightarrow H^{+} + HA^{-}}\) \(2nd \ dissociation: \mathrm{HA}^{-}\longleftrightarrow \mathrm{H^{+}+A^{2-}}\) Since we are mainly concerned with the first dissociation step, we will focus on the reaction involving \(\mathrm{H}_{2}\mathrm{A}\) and \(\mathrm{HA}^{-}\).
02

Set up the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \(pH = pK_a + \log \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\) We are given that the buffer should have a pH of 6.50 and the \(K_{a1} = 2 \times 10^{-2}\). Now, we need to find the amount of \(\mathrm{NaOH}\) to be added to form the pH 6.50 buffer. \(p_{K_{a1}} = -\log K_{a1}\) Calculate \(p_{K_{a1}}\): \(p_{K_{a1}} = -\log(2 \times 10^{-2}) \approx 1.70\)
03

Calculate the required ratio of [A-]/[HA]

Now, plug the pH and \(p_{K_{a1}}\) values into the Henderson-Hasselbalch equation and solve for the ratio \(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\): \(6.50 = 1.70 + \log \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\) Calculate the ratio of [A-]/[HA]: \(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 10^{(6.50-1.70)} \approx 10^{4.8} \approx 63.1\)
04

Relate the ratio to the amount of NaOH

The reaction between \(\mathrm{H}_{2}\mathrm{A}\) and \(\mathrm{NaOH}\) is: \(\mathrm{H}_{2}\mathrm{A+NaOH\to HA^{-}+H_2O+Na^{+}}\) Let x moles of \(\mathrm{NaOH}\) be added to 1.0 L of 1.0 M \(\mathrm{H}_{2}\mathrm{A}\). According to the reaction, x moles of \(\mathrm{HA}^{-}\) and x moles of \(\mathrm{Na^{+}}\) will be formed. After the reaction, [A-] = x M and [HA] = (1.0 M - x) M. Now, we can use the ratio calculated in step 3 to find x: \(\frac{x}{1-x} = 63.1\)
05

Calculate the moles of NaOH needed

Solve for x: \(x = \frac{63.1}{63.1 +1} = \frac{63.1}{64.1} \approx 0.984 M\) Since the volume of the acid solution is 1.0 L, the moles of NaOH required would be: Moles of NaOH = (Molarity of NaOH) x (Volume of the solution in liters) Moles of NaOH = 0.984 moles/L x 1.0 L = 0.984 moles
06

Find the volume of NaOH solution to be added

Recall that the \(\mathrm{NaOH}\) solution is at a concentration of 1.0 M. To find the volume of the \(\mathrm{NaOH}\) solution that needs to be added, divide the moles of \(\mathrm{NaOH}\) by its concentration: \(Volume \ of \ NaOH \ solution = \frac{0.984 \ moles \ NaOH}{1.0 \ M \ NaOH} = 0.984 \ L\) Therefore, to obtain a pH 6.50 buffer, you should add 0.984 L of the 1.0 M NaOH solution to 1.0 L of the 1.0 M H2A solution.

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Most popular questions from this chapter

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is $2.5 \times 10^{-14} .(\mathbf{a})$ What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to $1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$

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