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Chapter 17: Problem 70

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

Expert verified
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

01

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.
02

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.
03

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.

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Most popular questions from this chapter

What is the \(\mathrm{pH}\) of a solution obtained by adding \(1.15 \mathrm{mg}\) of aniline hydrochloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\right)\) to \(3.18 \mathrm{L}\) of \(0.105 \mathrm{M}\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right) ?\)

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Calculate the change in pH that results from adding (a) \(0.100 \mathrm{mol} \mathrm{NaNO}_{2}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq})\) (b) \(0.100 \mathrm{mol} \mathrm{NaNO}_{3}\) to \(1.00 \mathrm{L}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}(\mathrm{aq})\) Why are the changes not the same? Explain.

Two solutions are mixed: \(100.0 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) with \(\mathrm{pH} 2.50\) and \(100.0 \mathrm{mL}\) of \(\mathrm{NaOH}(\text { aq) with } \mathrm{pH} 11.00\) What is the pH of the resulting solution?

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