Chapter 15: Problem 87

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between \(7.1\) and \(7.2\). a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}{ }^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15\) ? \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

Short Answer

Expert verified

a. The concentration ratio of \(\mathrm{H}_2\mathrm{PO}_4^-\) to \(\mathrm{HPO}_4^{2-}\) in intracellular fluid at \(\mathrm{pH} = 7.15\) is: \[\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]} = 10^{(7.15 - pK_a)}\] b. The buffer composed of \(\mathrm{H}_3\mathrm{PO}_4\) and \(\mathrm{H}_2\mathrm{PO}_4^-\) is ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid because the \(pK_a\) value for the \(H_3PO_4\) reaction is significantly larger than the desired pH range, making it a poor buffer for maintaining the pH of intracellular fluid. Effective buffers have \(pK_a\) values close to the pH range to be maintained.

Step by step solution

The Henderson-Hasselbalch equation relates pH, the acid dissociation constant (\(K_a\)), and the concentrations of the acid and its conjugate base in a buffer solution. For this problem, the equation is given as: \[pH = pK_a + log \frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]}\] #Step 2: Find \(pK_a\) from given \(K_a\) values#

To calculate the concentration ratio, first, we need to find the \(pK_a\) values for the given dissociation constants \(K_a\). For \(H_2PO_4^- \rightleftharpoons HPO_4^{2-} + H^+\), \(K_a = 6.2 \times 10^{-8}\), so: \[pK_a = -log(6.2 \times 10^{-8})\] Similarly, for \(H_3PO_4 \rightleftharpoons H_2PO_4^- + H^+\), \(K_a = 7.5 \times 10^{-3}\), so: \[pK_a = -log(7.5 \times 10^{-3})\] #Step 3: Calculate the concentration ratio of \(H_2PO_4^-\) to \(HPO_4^{2-}\) at pH 7.15#

Here, the pH of intracellular fluids is given as 7.15, and we know the \(pK_a\) value calculated in step 2. We will now use the Henderson-Hasselbalch equation to find the concentration ratio: \[7.15 = pK_a + log \frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]}\] Solving for the ratio: \[\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]} = 10^{(7.15 - pK_a)}\] #Step 4: Explain why the buffer composed of \(H_3PO_4\) and \(H_2PO_4^-\) is ineffective#

The buffer composed of \(H_3PO_4\) and \(H_2PO_4^-\) would be ideal if the pK_a value were close to the pH at which we want to maintain the intracellular fluid (7.1-7.2). However, the calculated pK_a value for the \(H_3PO_4\) reaction (Step 2) is significantly larger than the desired pH range, making it a poor buffer in maintaining the pH of intracellular fluid. Effective buffers have pK_a values close to the pH range to be maintained.