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Chapter 20: Problem 56

Phosphate buffers are important in regulating the pH of intracellular fluids. If the concentration ratio of $\mathrm{H}_{2} \mathrm{PO}_{4}^{-} / \mathrm{HPO}_{4}^{2-}\( in a sample of intracellular fluid is \)1.1 : 1,$ what is the pH of this sample of intracullular fluid? $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$

Short Answer

Expert verified
Using the Henderson-Hasselbalch equation and the given concentration ratio of \(\mathrm{H_{2}PO_{4}^{-}}\) to \(\mathrm{HPO_{4}^{2-}}\) (1.1 : 1) and the \(K_\mathrm{a}\) value (\(6.2 \times 10^{-8}\)), the pH of the sample of intracellular fluid can be calculated as follows: $$ \mathrm{pH} = 7.21 + \log \frac{1.1}{1} \approx 7.34 $$ Hence, the pH of the sample of intracellular fluid is approximately 7.34.

Step by step solution

01

Identify the given values and write down the Henderson-Hasselbalch equation.

We are given the concentration ratio of \(\mathrm{H_{2}PO_{4}^{-}}\) to \(\mathrm{HPO}_{4}^{2-}\) and the \(K_\mathrm{a}\) value for the reaction. We will use the Henderson-Hasselbalch equation: $$ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} $$ where pH is the desired value, \(K_\mathrm{a}\) is the acid dissociation constant, [A⁻] is the concentration of the conjugate base (in this case, \(\mathrm{HPO_{4}^{2-}}\)), and [HA] is the concentration of the weak acid (in this case, \(\mathrm{H_{2}PO_{4}^{-}}\)).
02

Rewrite the concentration ratio as a fraction.

The given ratio is 1.1 : 1, which means there is 1.1 mole of \(\mathrm{H_{2}PO_{4}^{-}}\) for every 1 mole of \(\mathrm{HPO_{4}^{2-}}\). To use in the Henderson-Hasselbalch equation, rewrite it as a fraction: $$ \frac{[\mathrm{H_{2}PO_{4}^{-}}]}{[\mathrm{HPO_{4}^{2-}}]} = \frac{1.1}{1} $$
03

Convert the given Ka value to pKa.

Take the negative logarithm of the given Ka value (\(6.2 \times 10^{-8}\)) in order to find the pKa: $$ \mathrm{p}K_\mathrm{a} = -\log (6.2 \times 10^{-8}) \approx 7.21 $$
04

Plug values into the Henderson-Hasselbalch equation.

Now, we have all the values needed to use the Henderson-Hasselbalch equation and solve for the pH: $$ \mathrm{pH} =7.21+ \log \frac{1.1}{1} $$
05

Calculate the pH.

Perform the calculations in the equation: $$ \mathrm{pH} = 7.21 + \log (1.1) \approx 7.34 $$ The pH of the sample of intracellular fluid is approximately 7.34.

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