Question

Phosphate-Buffered Saline pH and Osmolarity Phosphate-buffered saline (PBS) is a solution commonly used in studies of animal tissues and cells. Its composition is 137 m? \(\mathrm{NaCl}, 2.7\) m? \(\mathrm{KCl}, 10 \mathrm{mM} \mathrm{Na}_{2} \mathrm{HPO}_{4}\left(\mathrm{pK}_{\mathrm{n}}=2.14\right), 1.8 \mathrm{~mm}\) \(\mathrm{KH}_{2} \mathrm{PO}_{4}\left(p K_{\mathrm{a}}=6.86\right)\). Calculate the pH and osmolarity of PBS. Give the osmolarity in units of osmoles per liter (osm/L).

Short answer

The pH is approximately 7.61, and the osmolarity is about 303 mOsm/L.

Step-by-step solution

Determine Contributions to pH

To find the pH of the solution, focus on the phosphate buffering components. The primary buffering species are the phosphate ions: \(\text{Na}_2\text{HPO}_4\) and \(\text{KH}_2\text{PO}_4\). The pH can be estimated using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]Here, the base is \(\text{Na}_2\text{HPO}_4\) and the acid is \(\text{KH}_2\text{PO}_4\). Given \(\text{pK}_a = 6.86\) for \(\text{KH}_2\text{PO}_4\), substitute the concentrations: \[ \text{pH} = 6.86 + \log\left(\frac{10}{1.8}\right) \]

Calculate the pH

Applying the logarithmic calculation:\[ \text{pH} = 6.86 + \log(5.56) \]Calculating:\[ \text{log}(5.56) \approx 0.746 \]So:\[ \text{pH} = 6.86 + 0.746 = 7.606 \]Thus, the pH of the PBS solution is approximately 7.61.

Calculate Osmolarity from Ionic Compounds

Osmolarity is calculated based on the total concentration of ions in the solution. Compute individual contributions from \(\text{NaCl}, \text{KCl}, \text{Na}_2\text{HPO}_4, \text{and } \text{KH}_2\text{PO}_4\):- \(\text{NaCl} = 137 \text{ mM}\) - \(\text{KCl} = 2.7 \text{ mM}\)- \(\text{Na}_2\text{HPO}_4 = 10 \text{ mM}\)- \(\text{KH}_2\text{PO}_4 = 1.8 \text{ mM}\)Osmolarity is the sum of all ionic species contributed by these salts.- From \(137 \text{ mM} \) \(\text{NaCl}\): \(\text{Na}^+ \text{ and Cl}^-\) contribute: \(2 \times 137 = 274 \text{ mOsm/L}\).- From \(2.7 \text{ mM} \) \(\text{KCl}\): \(\text{K}^+ \text{ and Cl}^-\) contribute: \(2 \times 2.7 = 5.4 \text{ mOsm/L}\).- From \(10 \text{ mM} \) \(\text{Na}_2\text{HPO}_4\), two \(\text{Na}^+\) and one \(\text{HPO}_4^{2-}\) contribute: \(2 \times 10 = 20 \text{ mOsm/L}\).- From \(1.8 \text{ mM} \) \(\text{KH}_2\text{PO}_4\), one \(\text{K}^+\) and one \(\text{H}_2\text{PO}_4^-\) contribute: \(2 \times 1.8 = 3.6 \text{ mOsm/L}\).

Calculate Total Osmolarity

Sum the contributions to find total osmolarity:\[ \text{Total Osmolarity} = 274 + 5.4 + 20 + 3.6 = 303 \text{ mOsm/L} \]Thus, the osmolarity of the phosphate-buffered saline is approximately \(303 \text{ mOsm/L}\).

Key concepts

pH Calculation

The pH of a solution tells us how acidic or basic it is. To figure out the pH of phosphate-buffered saline (PBS), we need to consider the main components that act as buffers. In PBS, the buffering action primarily comes from phosphate ions. Specifically, the compounds involved are \(\text{Na}_2\text{HPO}_4\) and \(\text{KH}_2\text{PO}_4\). To calculate the pH, we use the Henderson-Hasselbalch equation, which is a simple formula:\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]Here, the base is \(\text{Na}_2\text{HPO}_4\) and the acid is \(\text{KH}_2\text{PO}_4\). The given \(\text{pK}_a\) value for \(\text{KH}_2\text{PO}_4\) is 6.86. Using their concentrations, the pH is computed as follows:\[ \text{pH} = 6.86 + \log\left(\frac{10}{1.8}\right) \]By calculating the logarithm, we find:\[ \text{log}(5.56) \approx 0.746 \]So, the pH becomes:\[ \text{pH} = 6.86 + 0.746 = 7.606 \]This reveals that PBS has a slightly basic pH, around 7.61.

Osmolarity Calculation

Osmolarity tells us the total concentration of solute particles in a solution. It's a key measurement in solutions like phosphate-buffered saline (PBS). It ensures that cell environments stay balanced. To determine the osmolarity of PBS, we need to sum up the contributions from all dissolved ionic compounds.Several ions contribute to PBS's osmolarity:

• From \(137 \text{ mM} \) \(\text{NaCl}\): Each \(\text{NaCl}\) provides a \(\text{Na}^+\) and a \(\text{Cl}^-\), contributing \(2 \times 137 = 274 \text{ mOsm/L}\).
• From \(2.7 \text{ mM} \) \(\text{KCl}\): Each \(\text{KCl}\) gives a \(\text{K}^+\) and a \(\text{Cl}^-\), adding \(2 \times 2.7 = 5.4 \text{ mOsm/L}\).
• From \(10 \text{ mM} \) \(\text{Na}_2\text{HPO}_4\): Two \(\text{Na}^+\) and one \(\text{HPO}_4^{2-}\) ion result in \(2 \times 10 = 20 \text{ mOsm/L}\).
• From \(1.8 \text{ mM} \) \(\text{KH}_2\text{PO}_4\): A \(\text{K}^+\) and a \(\text{H}_2\text{PO}_4^-\) contribute \(2 \times 1.8 = 3.6 \text{ mOsm/L}\).

Adding these:\[ \text{Total Osmolarity} = 274 + 5.4 + 20 + 3.6 = 303 \text{ mOsm/L} \]This demonstrates that the osmolarity of PBS is about \(303 \text{ mOsm/L}\).

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a widely-used formula in chemistry and biology that links pH, \(\text{pK}_a\), and the concentration of acid and conjugate base in solutions. It provides a straightforward method to predict or adjust the pH of buffer systems, like phosphate-buffered saline (PBS).The formula looks like this:\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]Where:

• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the acid.
• \(\text{pK}_a\) represents the negative base-10 logarithm of the acid's dissociation constant.

This equation is crucial when adjusting the pH of solutions in lab settings, allowing predictions of how different concentrations of acid and base will change the pH. For instance, in PBS with buffering components \(\text{Na}_2\text{HPO}_4\) and \(\text{KH}_2\text{PO}_4\), the equation helps us predict the pH, as shown in our earlier calculation, making it invaluable for maintaining specific pH ranges needed in biological experiments.