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Chapter 20: Problem 50

Phosphate buffers are important in regulating the \(\mathrm{pH}\) 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 \(\mathrm{pH}\) of this sample of intracullular fluid? \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-\mathrm{s}}\)

Short Answer

Expert verified
The pH of the intracellular fluid in the given sample is approximately 7.17.

Step by step solution

01

Write the expression for Ka

Ka is the equilibrium constant for the acid dissociation reaction. In this case, H2PO4- is the acidic species and its Ka is given. The Ka expression for the given reaction is: Ka = \(\dfrac{[\mathrm{H}^{+}][\mathrm{HPO}_{4}^{2-}]}{[\mathrm{H}_{2}\mathrm{PO}_{4}^{-}]}\) Where Ka = 6.2 x 10^-8 (provided in the question)
02

Write the parameters in terms of variables

Given the ratio H2PO4- / HPO42- = 1.1 : 1, let the concentrations of H2PO4- and HPO42- be 1.1x and x, respectively. Let's also assume the concentration of hydrogen ions [H+] to be y. Now we can rewrite our expression for Ka using these variables: Ka = \(\dfrac{y \cdot x}{1.1x}\) Since Ka = 6.2 x 10^-8, now we can substitute the value 6.2 x 10^-8 = \(\dfrac{y \cdot x}{1.1x}\)
03

Simplify the equation and solve for [H+]

We can cancel out the x in the numerator and the denominator, and then solve for y: 6.2 x 10^-8 = \(\dfrac{y}{1.1}\) Multiplying both sides by 1.1, we get: y = 6.2 x 10^-8 * 1.1 y = 6.82 x 10^-8
04

Calculate the pH of the solution

Now that we have the concentration of hydrogen ions [H+], we can calculate the pH using the following formula: pH = -log10([H+]) pH = -log10(6.82 x 10^-8) After evaluating the expression, we get: pH ≈ 7.17 So, the pH of this sample of intracellular fluid is approximately 7.17.

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