Question 16.SIE.1: Phosphorous acid (H3PO3) has the Lewis structure shown at ri...
Phosphorous acid (H_3PO_3) has the Lewis structure shown at right.
(a) Explain why H_3PO_3 is diprotic and not triprotic. (b) A 25.0 mL sample of an H_3PO_3 solution titrated with 0.102 M NaOH requires 23.3 mL of NaOH to neutralize both acidic protons. What is the molarity of the H_3PO_3 solution? (c) The original solution from part (b) has a pH of 1.59. Calculate the percent ionization and K_{a1} for H_3PO_3, assuming that K_{a 1}\gg K_{a 2}. (d) How does the osmotic pressure of a 0.050 M solution of HCl compare qualitatively with that of a 0.050 M solution of H_3PO_3? Explain.
We will use what we have learned about molecular structure and its impact on acidic behavior to answer part (a). We will then use stoichiometry and the relationship between pH and [H^+] to answer parts (b) and (c). Finally, we will consider percent ionization in order to compare the osmotic pressure of the two solutions in part (d).
(a) Acids have polar H—X bonds. From Figure 8.8 (p. 383) we see that the electronegativity of H is 2.1 and that of P is also 2.1. Because the two elements have the same electronegativity, the H—P bond is nonpolar. Thus, this H cannot be acidic. The other two H atoms, however, are bonded to O, which has an electronegativity of 3.5. The H—O bonds are, therefore, polar with H having a partial positive charge. These two H atoms are consequently acidic.
(b) The chemical equation for the neutralization reaction is:
H _3 PO _3(a q)+2 NaOH (a q) \longrightarrow Na _2 HPO _3(a q)+2 H _2 O (l)
From the definition of molarity, M = mol/L, we see that moles = M × L. Thus, the number of moles of NaOH added to the solution is:
(0.0233 \,\cancel{L} )(0.102 \,mol / \cancel{L} )=2.38 \times 10^{-3}\,mol \,NaOH
The balanced equation indicates that 2 mol of NaOH is consumed for each mole of H_3PO_3. Thus, the number of moles of H_3PO_3 in the sample is:
\left(2.38 \times 10^{-3}\,\cancel{mol \,NaOH} \right)\left(\frac{1 \,mol \,H _3 PO _3}{2 \,\cancel{mol \,NaOH}}\right)=1.19 \times 10^{-3}\,mol \,H _3 PO _3
The concentration of the H_3PO_3 solution, therefore, equals \left(1.19 \times 10^{-3}\,mol \right) /(0.0250 \,L )=0.0476\, M
(c) From the pH of the solution, 1.59, we can calculate [H^+] at equilibrium:
\left[ H ^{+}\right]=\operatorname{antilog}(-1.59)=10^{-1.59}=0.026 \,M \,\text{(two significant figures)}
Because K_{a 1}\gg K_{a 2}, the vast majority of the ions in solution are from the first ionization step of the acid.
Because one H_2PO_3^- ion forms for each H^+ ion formed, the equilibrium concentrations of H^+ and H_2PO_3^- are equal:\left[ H ^{+}\right]=\left[ H _2 PO _3^{-}\right]=0.026 \,M. The equilibrium concentration of H_3PO_3 equals the initial concentration minus the amount that ionizes to form H^+ and H _2 PO _3^{-}:\left[ H _3 PO _3\right]=0.0476 \,M -0.026\, M=0.022 \,M (two significant figures). These results can be tabulated as follows:
\begin{array}{|l|c|c|c|}\hline&H _3 PO _3(a q) \rightleftharpoons&H ^{+}(a q)+&H _2 PO _3^{-}(a q)\\\hline \begin{array}{l}\text{Initial}\\\text{concentration}(M)\end{array}& 0.0476 & 0 & 0 \\\hline \begin{array}{l}\text{Change in}\\\text{concentration}(M)\end{array}& -0.026 & +0.026 & +0.026 \\\hline \begin{array}{l}\text{Equilibrium}\\\text{concentration}(M)\end{array}& 0.022 & 0.026 & 0.026 \\\hline\end{array}
The percent ionization is:
\text{percent ionization}=\frac{\left[ H ^{+}\right]_{\text{equilibrium}}}{\left[ H _3 PO _3\right]_{\text{initial}}}\times 100 \%=\frac{0.026\,\cancel{M}}{0.0476 \,\cancel{M}}\times 100 \%=55 \%
The first acid-dissociation constant is:
K_{a 1}=\frac{\left[ H ^{+}\right]\left[ H _2 PO _3^{-}\right]}{\left[ H _3 PO _3\right]}=\frac{(0.026)(0.026)}{0.022}=0.031
(d) Osmotic pressure is a colligative property and depends on the total concentration of particles in solution. Because HCl is a strong acid, a 0.050 M solution will contain 0.050 M H^+(aq) and 0.050 M Cl^-(aq),or a total of 0.100 mol/L of particles. Because H_3PO_3 is a weak acid, it ionizes to a lesser extent than HCl and, hence, there are fewer particles in the H_3PO_3 solution. As a result, the H_3PO_3 solution will have the lower osmotic pressure.
