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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

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

Step by step solution

01

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: \[ \text{pH} = 6.86 + \log\left(\frac{10}{1.8}\right) \]
02

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.
03

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: \(2 \times 1.8 = 3.6 \text{ mOsm/L}\).
04

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 \(303 \text{ mOsm/L}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

a. In what pH range can glycine be used as an effective buffer due to its amino group? b. In a \(0.1 \mathrm{~m}\) solution of glycine at pH \(9.0\), what fraction of glycine has its amino group in the \(-\mathrm{NH}_{3}^{4}\) form? c. How much \(5 \mathrm{M}\) KOH must be added to \(1.0 \mathrm{~L}\) of \(0.1 \mathrm{M}\) glycine at pH \(9.0\) to bring its pII to exactly \(10.0 ?\) d. When 9996 of the glycine is in ?ts \(-\mathrm{NH}_{3}^{+}\)form, what is the numerical relation between the pH of the solution and the p \(K_{\mathrm{n}}\) of the amino group? Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a \(\mathrm{p} K_{\mathrm{n}}\) of \(9.6\), can exist either in the protonated form \(\left(-\mathrm{NH}_{3}^{+}\right)\)or as the free base \(\left(-\mathrm{NH}_{2}\right)\), because of the reversible equilibrium $$ \mathrm{F}-\mathrm{NH}_{3}^{+} \rightleftharpoons \mathrm{H}-\mathrm{NH}_{2}+\mathrm{H}^{+} $$

Preparation of a Phosphate Buffer Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\), a triprotic acid, has three \(\mathrm{p} K_{\mathrm{a}}\) values: \(2.14,6.86\), and 12.4. What molar ratio of \(\mathrm{HPO}_{4}^{2-}\) to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)in solution would produce a \(\mathrm{pH}\) of \(7.0 ?\) Hint: Only one of the \(\mathrm{p} K_{\mathrm{a}}\) values is relevant here.

Determining Charge and Solubility of Organic Acids Suppose that, for a typical carboxyl-containing compound, the \(\mathrm{p} K_{\mathrm{a}}\) is approximately 3. Suppose \(\mathrm{HOOC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COOH}, \mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COOH}_{2}\) and \(\mathrm{HOOC}-\left(\mathrm{CH}_{2}\right)_{2}-\mathrm{COOH}\) are added to water at pH \(7 .\) a. What is the net charge of each compound in the solution? b. List the compounds in order from most soluble to least soluble.

Calculation of Molar Ratios of Conjugate Base to Weak Acid from pll For a weak acid with a pK of \(6.00\), calculate the ratio of conjugate base to acid at a pH of \(5.00\).

Duration of Hydrogen Bonds PCR is a laboratory process in which specific DNA sequences are copied and amplified manyfold. The two DNA strands, which are held together in part by hydrogen bonds between them, are heated in a buffered solution to separate the two strands, then cooled to allow them to reassociate. What do you predict about the average duration of \(\mathrm{H}\) bonds at the high temperature in comparison to the low temperature?

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