Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intra- cellular fluids at pH values generally between 7.1 and \(7.2 .\) a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) inintracellular 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 pH of intracellular fluid? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$

To find the concentration ratio, we need to use the Henderson-Hasselbalch equation. The equation is as follows: \[ pH = pK_a + \log{\left(\frac{[A^-]}{[HA]}\right)} \] Where: - pH is the pH of the solution, - pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka) of the acid, - [A^-] is the concentration of the conjugate base, and - [HA] is the concentration of the weak acid. The given pH is 7.15 and the Ka value is \(6.2 \times 10^{-8}\). Therefore, the pK_a value is defined as: \[ pK_a = \log{\left(\frac{1}{Ka}\right)} = -\log{(6.2 \times 10^{-8})} \] Calculate the pKa value:

Using the given Ka value, we can calculate pKa as follows: \[ pK_a = -\log{(6.2 \times 10^{-8})} \approx 7.21 \] Now we can use the pH value of 7.15 and pKa value of 7.21 to find the concentration ratio of \(H_2PO_4^{-}\) to \(HPO_4^{2-}\) by substituting these values into the Henderson-Hasselbalch equation.

Now we can use the pH, 7.15, and the pKa, 7.21, in the Henderson-Hasselbalch equation to find the concentration ratio of \(H_2PO_4^-\) to \(HPO_4^{2-}\): \[ 7.15 = 7.21 + \log{\left(\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}\right)} \] Rearrange the equation to determine the concentration ratio: \[ \log{\left(\frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]}\right)} = 7.15 - 7.21 = -0.06 \] \[ \frac{[HPO_4^{2-}]}{[H_2PO_4^{-}]} = 10^{-0.06} \approx 0.861 \] Thus, the concentration ratio of \(H_2PO_4^-\) to \(HPO_4^{2-}\) in intracellular fluid at pH 7.15 is approximately \(0.861:1\).

In the second part of the exercise, we are given that: \[ H_3PO_4 \leftrightharpoons H_2PO_4^- + H^+ \quad K_2 = 7.5 \times 10^{-3} \] Now we need to calculate the pK_a2 value to determine whether it falls within the desired pH range for intracellular fluid buffering: \[ pK_a2 = -\log{(7.5 \times 10^{-3})} \] Then: \[ pK_a2 = -\log{(7.5 \times 10^{-3})} \approx 2.12 \] Now let's compare the pKa values of both reactions: For the phosphate buffer: \[ pK_a = -\log{(6.2 \times 10^{-8})} = 7.21 \] For the \(H_3PO_4\) buffer: \[ pK_a2 = -\log{(7.5 \times 10^{-3})} = 2.12 \] The pKa of the \(H_3PO_4\) buffer reaction is 2.12, which is significantly lower than the desired pH range of intracellular fluid (7.1 and 7.2). Thus, a buffer composed of \(H_3PO_4\) and \(H_2PO_4^-\) is ineffective in buffering the pH of intracellular fluid as the buffering capacity of the \(H_3PO_4\) buffer reacts more strongly with added acidic or basic compounds, leading to a larger change in pH. The efficiency of a buffer is optimal when the pKa is near the pH of the system being buffered.