Question

You are asked to prepare a \(\mathrm{KH}_{2} \mathrm{PO}_{4}-\mathrm{Na}_{2} \mathrm{HPO}_{4}\) solu- tion that has the same \(\mathrm{pH}\) as human blood, 7.40 (a) What should be the ratio of concentrations \(\left[\mathrm{HPO}_{4}^{2-}\right] /\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]\) in this solution? (b) Suppose you have to prepare \(1.00 \mathrm{L}\) of the solution described in part (a) and that this solution must be isotonic with blood (have the same osmotic pressure as blood). What masses of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and of \(\mathrm{Na}_{2} \mathrm{HPO}_{4} \cdot 12 \mathrm{H}_{2} \mathrm{O}\) would you use? [Hint: Refer to the definition of isotonic on page \(580 .\) Recall that a solution of \(\mathrm{NaCl}\) with \(9.2 \mathrm{g} \mathrm{NaCl} / \mathrm{L}\) solution is isotonic with blood, and assume that \(\mathrm{NaCl}\) is completely ionized in aqueous solution.]

Short answer

The ratio of concentrations of \(\mathrm{HPO}_{4}^{2-} / \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) should be approximately 1.58. To prepare 1.00 L of the isotonic solution, use approximately 4.94 g of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and approximately 22.8 g of \(\mathrm{Na}_{2} \mathrm{HPO}_{4} . 12 \mathrm{H}_{2} \mathrm{O}\).

Step-by-step solution

Calculate the ratio of concentrations

The pH of a solution of a weak acid and its conjugate base can be calculated using the Henderson-Hasselbalch equation: \[pH = pKa + log_{10} \left(\frac{\([bases]}\){[acid]}\right)\]Since the pKa for H2PO4- is 7.2 and the pH required is 7.4, let's solve for the ratio of [HPO4^2-]/[H2PO4^-]\[7.4 = 7.2 + log_{10} \left(\frac{\([HPO4^{2-}]\)}{[H2PO4^{-}]}\right)\]Solving this gives the ratio of [HPO4^2-]/[H2PO4^-] as approximately 1.58.

Calculate the masses for preparing isotonic solution

For isotonicity with blood, the overall molarity of ions should be similar to that of blood (0.15 M). A solution of NaCl with 9.2 g NaCl / L solution is isotonic with blood, and \(\mathrm{NaCl}\) is completely ionized in aqueous solution, providing 0.15 M of Na+ and 0.15 M of Cl-. Therefore, the total moles of ionic species in the prepared solution should be approximately 0.30 moles per litre. As both \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and \(\mathrm{Na}_{2} \mathrm{HPO}_{4} . 12 \mathrm{H}_{2} \mathrm{O}\) each ionize to give 3 moles of ions per mole of salt, the sum of moles of KH2PO4 and Na2HPO4 should be 0.30/3 = 0.10 moles. Using the molar formula masses for \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) (136.09 g/mol) and \(\mathrm{Na}_{2} \mathrm{HPO}_{4} . 12 \mathrm{H}_{2} \mathrm{O}\) (358.14 g/mol), and maintaining the previously calculated ratio, we can calculate the required masses of each.

Compute the required masses of each

Since the required ratio is 1.58 we can find the proportions of each as x and 1.58x which must sum to give 0.10. Solving for x gives approximately 0.0363. Therefore, the required mass of KH2PO4 is approximately 0.0363 mol x 136.1 g/mol = 4.94 g and the required mass of Na2HPO4.12H2O is approximately 0.0637 mol x 358.1 g/mol = 22.8 g.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a very useful formula in chemistry, especially when dealing with buffer solutions. It allows us to relate the pH of a solution to the pKa, and the ratio of concentrations of a conjugate base and its corresponding acid.
This formula can be expressed as:\[\text{pH} = \text{pKa} + \log_{10} \left(\frac{[\text{base}]}{[\text{acid}]}\right)\]Here, "pH" is what we want to find or adjust to a specific value, like the blood pH of 7.4. "pKa" is the negative logarithm of the acid dissociation constant, a measure of the strength of the acid. The fraction is the ratio of the concentrations of the conjugate base to the acid. Using this equation becomes essential when you need to make a buffer solution with a specific pH.
In our exercise, the requirement to achieve the pH of 7.4 necessitated calculating the ratio of the concentrations for the phosphate buffer solution: \[\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^{-}]}\]By analyzing the equation and the given pKa value of approximately 7.2, the ratio determined was 1.58. This means you need 1.58 times more of the conjugate base \(\text{HPO}_4^{2-}\) compared to the acid \(\text{H}_2\text{PO}_4^{-}\) to achieve the desired pH.

pH calculation

Calculating the pH of a buffer involves understanding a few core concepts. First, recognize the nature of the solution — is it acidic or basic? This affects how you approach the calculations and decisions on chemical additions.
To calculate the pH, you can follow these simple steps:

• Determine the identities of the acid and its conjugate base in the buffer system. For instance, \(\text{KH}_2\text{PO}_4\) is the acid and \(\text{Na}_2\text{HPO}_4\) is the conjugate base.
• Utilize the Henderson-Hasselbalch equation, which helps predict the pH based on the known pKa of the acid involved.
• Adjust concentrations as necessary to achieve the correct pH. In this case, achieving a pH of 7.4 required balancing the concentrations according to the calculated ratio of about 1.58.

These steps ensure that the mixture will resist changes in pH, even if small amounts of acid or base are introduced, thereby efficiently holding the desired pH around 7.4.

Isotonic solutions

Isotonic solutions are critical, especially in medical and biological contexts, where solutions need to be compatible with bodily fluids to avoid causing harm.
In simple terms, an isotonic solution is one that has the same osmotic pressure as another solution, typically physiological fluids like blood. This means there's an equal concentration of solutes inside and outside cells, preventing the movement of water into or out of the cells.
In the context of our exercise:

• It required a solution that had the same osmotic pressure as blood. Hence, mimicking the molarity of ions, which is approximately 0.30 moles per litre due to contributions from both the sodium and chloride ions in blood, typically assumed with a NaCl solution.
• To achieve this with \(\text{KH}_2\text{PO}_4\) and \(\text{Na}_2\text{HPO}_4\), one must calculate the total molarity of ions from the dissociation of these salts.
• This ensures that when composing the final solution, it will be isotonic, thereby safe for applications involving human physiology.

This isotonic condition is crucial for safe interactions with cells and tissues, maintaining homeostasis without causing cellular distress. By carrying out the detailed molarity calculations, the solution's ionic strength was matched to that of blood.