Problem 50 - Chemistry An Atoms First Approach

Question

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 \(p\) H of this sample of intracellular fluid? \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\).

Step-by-Step Solution

Answer

The pH of this sample of intracellular fluid is approximately 7.21.

Step 1: Write down the given ratio

We are given the ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) as \(1.1:1\).

Step 2: Express the concentrations in terms of a variable

Let the concentration of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) be \(1.1x\) and the concentration of \(\mathrm{HPO}_{4}^{2-}\) be \(x\). Then the ratio will also be \(1.1 : 1\).

Step 3: Write the \(K_{a}\) expression

The expression for \(K_{a}\) is the concentration of products divided by the concentration of reactants: \[K_{a} = \frac{[\mathrm{HPO}_{4}^{2-}] [\mathrm{H}^{+}]}{[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}]}\]

Step 4: Substitute the given values into the \(K_{a}\) expression

Substitute the values of \([\mathrm{H}_{2} \mathrm{PO}_{4}^{-}]\), \([\mathrm{HPO}_{4}^{2-}]\), and \(K_{a}\) into the expression: \[6.2 \times 10^{-8} = \frac{x \cdot [\mathrm{H}^{+}]}{1.1x}\]

Step 5: Solve for \([\mathrm{H}^{+}]\)

We can now simplify the equation and solve for \([\mathrm{H}^{+}]\): \[6.2 \times 10^{-8} \times 1.1x = x \cdot [\mathrm{H}^{+}]\] \[[\mathrm{H}^{+}] = 6.2 \times 10^{-8} \times 1.1\]

Step 6: Calculate the pH

Now that we have the concentration of \([\mathrm{H}^{+}]\), we can use the pH formula: \[\text{pH} = -\log{[\mathrm{H}^{+}]}\] Plug in the value of \([\mathrm{H}^{+}]\): \[\text{pH} = -\log(6.2 \times 10^{-8} \times 1.1)\]

Step 7: Calculate the final pH value

Using a calculator, compute the pH value: \[\text{pH} \approx 7.21\] So, the pH of this sample of intracellular fluid is approximately 7.21.

Key Concepts

Buffer Systems

Buffer systems are crucial in maintaining the stability of the pH levels in various environments, such as biological fluids, by minimizing changes upon the addition of an acid or a base. A buffer consists of a weak acid and its conjugate base or a weak base and its conjugate acid. In the body, one of the most important buffer systems is the phosphate buffer. It contains the dihydrogen phosphate ion, \(\mathrm{H}_2\mathrm{PO}_4^{-}\), and hydrogen phosphate ion, \(\mathrm{HPO}_4^{2-}\). The unique ability of this system to resist pH changes makes it fundamental for processes in the cells where a specific and stable pH is necessary for enzyme activity and other biochemical reactions.

When an acid (proton donor) is added to the system, the base component \(\mathrm{HPO}_4^{2-}\) neutralizes it by accepting protons, forming more \(\mathrm{H}_2\mathrm{PO}_4^{-}\). Conversely, when a base (proton acceptor) is added, the acid component \(\mathrm{H}_2\mathrm{PO}_4^{-}\) donates protons to form more \(\mathrm{HPO}_4^{2-}\), thereby preventing significant changes in pH. Understanding buffer systems is essential for students troubleshooting pH-related exercises, as it provides the framework for predicting the behavior of solutions in acid-base reactions.

Acid-Base Equilibrium

Acid-base equilibrium is the balance between the concentrations of acids and bases in a solution. At this equilibrium, the rates at which the acid donates protons (H+) is equal to the rate at which the conjugate base accepts them. The expression for the equilibrium constant, \(K_{a}\), reflects the strength of an acid, proton donation propensity, and its ability to maintain the equilibrium when faced with external pH changes.

Considering the phosphate buffer from our exercise, the acid-base equilibrium involves the weak acid \(\mathrm{H}_2\mathrm{PO}_4^{-}\) and its conjugate base \(\mathrm{HPO}_4^{2-}\). The equilibrium constantly shifts to counter any disturbances as a result of Le Chatelier's principle, which is fundamental in understanding how buffers work to keep the pH relatively constant. A deep comprehension of this balance is necessary since it forms the basis of calculating the pH in buffer solutions.

pH Calculation

Calculating the pH is a fundamental aspect of chemistry that involves finding the acidity or basicity of a solution. The pH scale ranges from 0 to 14, where 7 is neutral, below 7 is acidic, and above 7 is basic. The pH is logarithmically based and calculated using the formula: \(\text{pH} = -\log[\mathrm{H}^{+}]\), where \([\mathrm{H}^{+}]\) is the concentration of hydrogen ions in moles per liter.

In the provided exercise, we find the pH by first understanding the ratio of the components in the phosphate buffer and then solving for the hydrogen ion concentration using the acid dissociation constant, \(K_{a}\). Once \([\mathrm{H}^{+}]\) is known, applying the pH formula yields the solution’s pH. It's important for students to be comfortable with logarithmic calculations and the pH scale to confidently solve these types of problems.

Ka Expression

The \(K_{a}\) expression, known as the acid dissociation constant, quantifies the strength of an acid in solution. It is defined by the equilibrium concentrations of the components of the acid and its conjugate base. The expression for \(K_{a}\) is: \[K_{a} = \frac{[\mathrm{A}^{-}] [\mathrm{H}^{+}]}{[\mathrm{HA}]},\] where \([\mathrm{HA}]\) is the concentration of the acid, \([\mathrm{A}^{-}]\) is the concentration of the conjugate base, and \([\mathrm{H}^{+}]\) is the concentration of hydrogen ions. Stronger acids have higher \(K_{a}\) values, indicating they donate protons more readily.

In the context of our exercise, we're dealing with a weak acid \(\mathrm{H}_2\mathrm{PO}_4^{-}\), which partially dissociates in solution. By rearranging the \(K_{a}\) expression, we were able to solve for \([\mathrm{H}^{+}]\) using the given \(K_{a}\) value and the ratio of acid to base concentrations. This understanding of the \(K_{a}\) expression is critical for students as it bridges the gap between theoretical concepts of acid strength and practical pH calculations.