Problem 7 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The pH inside most cells is maintained at around 7.4 by a phosphate buffer made up from the \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})\) ion and its conjugate base, \(\mathrm{HPO}_{4}^{2-}(\mathrm{aq})\). The \(\mathrm{pK}_{\mathrm{a}}\) of the acid \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})\) is 7.2. (Section 7.3) (a) Write an expression for \(K_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (aq). (b) Calculate the ratio of \(\left[\mathrm{HPO}_{4}^{2-}(\mathrm{aq})\right] /\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})\right]\) needed to give a pH of 7.4 in the cell. (c) A typical value for the total phosphate concentration in a cell, \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})\right]+\left[\mathrm{HPO}_{4}^{2-}(\mathrm{aq})\right],\) is \(0.020 \mathrm{moldm}^{-3}\) Calculate typical values of \(\left[\mathrm{HPO}_{4}^{2-}(\mathrm{aq})\right]\) and \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})\right]\) inside a cell.

Step-by-Step Solution

Answer

(a) \( K_a = \frac{[\mathrm{H}^+][\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]} \). (b) Ratio is approximately 1.58. (c) \([\mathrm{H}_2 \mathrm{PO}_4^-] = 0.00775\, \text{moldm}^{-3}\) and \([\mathrm{HPO}_4^{2-}] = 0.01225\, \text{moldm}^{-3}\).

Step 1: Expression for Ka

The expression for the acid dissociation constant, \( K_a \), of \( \mathrm{H}_2 \mathrm{PO}_4^- \) is based on the equilibrium reaction: \[ \mathrm{H}_2 \mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \] This gives us the expression: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]} \]

Step 2: Using the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \text{pK}_a + \log\left(\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]}\right) \] Substituting \( \text{pH} = 7.4 \) and \( \text{pK}_a = 7.2 \) yields: \[ 7.4 = 7.2 + \log\left(\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]}\right) \]

Step 3: Solving for the Ratio

Rearrange the equation to solve for \( \log\left(\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]}\right) \): \[ \log\left(\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]}\right) = 7.4 - 7.2 = 0.2 \] Therefore, the ratio is given by the anti-logarithm: \[ \frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2 \mathrm{PO}_4^-]} = 10^{0.2} \approx 1.58 \]

Step 4: Calculating Concentrations

Given \( [\mathrm{H}_2 \mathrm{PO}_4^-] + [\mathrm{HPO}_4^{2-}] = 0.020 \text{ moldm}^{-3} \) and the ratio found previously, let \( [\mathrm{H}_2 \mathrm{PO}_4^-] = x \) and \( [\mathrm{HPO}_4^{2-}] = 1.58x \). From the total concentration we have: \[ x + 1.58x = 0.020 \] Solving for \( x \) results in: \[ 2.58x = 0.020 \quad \Rightarrow \quad x = \frac{0.020}{2.58} \approx 0.00775 \text{ moldm}^{-3} \] This means \( [\mathrm{H}_2 \mathrm{PO}_4^-] = 0.00775 \text{ moldm}^{-3} \) and \( [\mathrm{HPO}_4^{2-}] = 1.58 \times 0.00775 \approx 0.01225 \text{ moldm}^{-3} \).

Key Concepts

pH regulation

In biological systems, maintaining a consistent pH is crucial for proper cellular function. Cells typically use buffer systems to regulate their pH levels, and one significant buffer system involves phosphate ions. A phosphate buffer consists of a weak acid, like dihydrogen phosphate (\( \mathrm{H}_2\mathrm{PO}_4^- \)), and its conjugate base, hydrogen phosphate (\( \mathrm{HPO}_4^{2-} \)). This buffer works to keep the internal pH around 7.4, optimal for many cellular processes.

The buffer works by neutralizing added acids or bases without significantly altering the overall pH. When an acid is introduced to the system, the excess hydrogen ions are absorbed by the conjugate base (\( \mathrm{HPO}_4^{2-} \)). Conversely, when a base is added, it can remove hydrogen ions from the weak acid (\( \mathrm{H}_2\mathrm{PO}_4^- \)), effectively counteracting the change.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation provides a way to calculate the pH of a solution using the concentration of acid and its conjugate base. The mathematical form of the equation is:\[\text{pH} = \text{pK}_a + \log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\]

This equation is especially useful in understanding how buffers work to stabilize pH. In the context of a phosphate buffer, it allows us to calculate the ratio of \( [\mathrm{HPO}_4^{2-}] \) to \( [\mathrm{H}_2\mathrm{PO}_4^-] \) required to achieve a desired pH, such as 7.4 in cells. In this specific situation, given \( \text{pK}_a = 7.2 \), a \( \text{pH} \) of 7.4 indicates that the system will favor slightly more conjugate base to weak acid, leading to a ratio of approximately 1.58.

acid dissociation constant

The acid dissociation constant, denoted as \( K_a \), quantifies the strength of an acid in solution. It refers to the equilibrium constant for the dissociation of acid into its ionized forms. For the acid \( \mathrm{H}_2\mathrm{PO}_4^- \), the dissociation into hydrogen ions and its conjugate base can be expressed as follows:\[\mathrm{H}_2\mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-}\]

The expression for \( K_a \) is therefore:\[K_a = \frac{[\mathrm{H}^+][\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]}\]

A higher \( K_a \) indicates a stronger acid, which dissociates more completely in water. For \( \mathrm{H}_2\mathrm{PO}_4^- \), its moderate \( K_a \) helps the phosphate buffer to efficiently resist changes in pH within the cellular environment.

equilibrium reactions

Equilibrium reactions are pivotal in maintaining balance in chemical systems. They occur when the rates of the forward and reverse reactions are equal, leading to stable concentrations of reactants and products over time.

In the case of the phosphate buffer within cells, the equilibrium can be seen in the conversion between \( \mathrm{H}_2\mathrm{PO}_4^- \) and \( \mathrm{HPO}_4^{2-} \). The equation involved is:\[\mathrm{H}_2\mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-}\]

At equilibrium, this reaction allows for the buffering action that stabilizes the pH. If there is an excess of \( \mathrm{H}^+ \), the equilibrium shifts to the left, decreasing \( \mathrm{H}^+ \) concentration, thus maintaining pH levels. Similarly, if \( \mathrm{H}^+ \) is depleted, the equilibrium shifts to the right, releasing \( \mathrm{H}^+ \) ions back into the solution. This dynamic balance is crucial for cell survival and function.