[latex]{K}_{c}=\dfrac{{\left[\text{C}\right]}^{2}}{\left[\text{A}\right]{\left[\text{B}\right]}^{2}}\text{.}[/latex] There are many different sets of equilibrium concentrations; two are [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.

1. The reaction may be written as

[latex]{\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\rightleftharpoons2{\text{NH}}_{3}\left(g\right)[/latex]

The equilibrium constant for the reaction is

[latex]{K}_{c}=\frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}=\frac{{\left(4.12\times {10}^{-1}\right)}^{2}}{\left(1.15\right){\left(1.35\right)}^{3}}=\frac{{\left(0.170\right)}^{2}}{\left(1.15\right){\left(2.46\right)}^{3}}0.0600=6.00\times10^{-2}[/latex]

3. The decomposition of PCl₅ to PCl₃ and Cl₂ is given as

[latex]\begin{array}{l}{\text{PCl}}_{5}\left(g\right)\rightleftharpoons{\text{PCl}}_{3}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\hfill \\{K}_{c}=\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}\hfill \end{array}[/latex]

Let x = change in [PCl₅].

[latex]{K}_{c}=\frac{\left(0.40\right)\left(0.40\right)}{\left(0.32\right)}=0.50[/latex]

5.  The equilibrium equation is

[latex]{K}_{P}=\frac{{\left[\text{NOCl}\right]}^{2}}{{\left[\text{NO}\right]}^{2}\left[{\text{Cl}}_{2}\right]}=\frac{{\left(1.2\right)}^{2}}{{\left(0.050\right)}^{2}\left(0.30\right)}=\frac{1.44}{\left(2.5\times {10}^{-3}\right)\left(0.30\right)}=1.9\times {10}^{3}[/latex]

7. Because the decomposition must generate the same pressure of HCl as NH₃, 1.75 atm of HCl must be present.

[latex]{K}_{p}={P}_{{\text{NH}}_{3}}{P}_{\text{HCl}}=\left(1.75\text{atm}\right)\left(1.75\text{atm}\right)=3.06[/latex]

1.  Write the equilibrium constant expression and solve for [NH₃].

[latex]{K}_{c}=\frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[{\text{N}}_{2}\right]{\left[{\text{H}}_{2}\right]}^{3}}=\frac{{\left[{\text{NH}}_{3}\right]}^{2}}{\left[1.2\right]{\left[0.24\right]}^{3}}=0.50[/latex]

[NH₃]² = 1.2 × (0.24)³ × 0.50 = 0.0083

[NH₃] = 9.1 × 10^-2M

3. Write the equilibrium constant expression and solve for [CO].

[latex]\begin{array}{l}{K}_{c}=\frac{\left[{\text{CO}}_{2}\right]}{\left[\text{CO}\right]}=\frac{0.100}{\left[\text{CO}\right]}=4.90\times {10}^{2}\hfill \\ \left[\text{CO}\right]=\frac{0.100}{4.90\times {10}^{2}}=2.0\times {0}^{-4}M\hfill \end{array}[/latex]

5. Calculate Q based on the calculated concentrations and see if it is equal to K_c. Because Q does equal 4.32, the system must be at equilibrium.

1. The changes in concentrations (or pressure, if requested) are as follows:

1. [latex]\begin{array}{llll}2{\text{SO}}_{3}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{SO}}_{2}\left(g\right)+\hfill & {\text{O}}_{2}\left(g\right)\hfill \\ \text{ }\hfill -2x & &2x\hfill & +x\hfill \\ -0.250M\hfill & & 0.250M\hfill & 0.125M\hfill \end{array}[/latex]
2. [latex]\begin{array}{lllll}4{\text{NH}}_{3}\left(g\right)\hfill & +3{\text{O}}_{2}\left(g\right)\hfill & \rightleftharpoons\hfill & 2{\text{N}}_{2}\left(g\right)+\hfill & 6{\text{H}}_{2}\text{O}\left(g\right)\hfill \\ 4x\hfill & 3x\hfill & & -2x\hfill & -6x\hfill \\ 0.32M\hfill & 0.24M\hfill & &-0.16M\hfill & -0.48M\hfill \end{array}[/latex]
3. Change in pressure:
   [latex]\begin{array}{llll}2{\text{CH}}_{4}\left(g\right)\hfill & \rightleftharpoons\hfill & {\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ -2x\hfill & & x\hfill & 3x\hfill \\ -50\text{ torr}\hfill & & 25\text{ torr}\hfill & 75\text{ torr}\hfill \end{array}[/latex]
4. Change in pressure:
   [latex]\begin{array}{lllll}{\text{CH}}_{4}\left(g\right)+\hfill & {\text{H}}_{2}\text{O}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{CO}\left(g\right)+\hfill & 3{\text{H}}_{2}\left(g\right)\hfill \\ x\hfill & x\hfill & & -x\hfill & -3x\hfill \\ 5\text{ atm}\hfill & 5\text{atm}\hfill & & -5\text{ atm}\hfill & -15\text{ atm}\hfill \end{array}[/latex]
5. [latex]\begin{array}{llll}{\text{NH}}_{4}\text{Cl}\left(s\right)\hfill & \rightleftharpoons\hfill & {\text{NH}}_{3}\left(g\right)+\hfill & \text{HCl}\left(g\right)\hfill \\ & & x\hfill & x\hfill \\ & & \hfill 1.03\times {10}^{-4}M\hfill & 1.03\times10^{-4}M\hfill \end{array}[/latex]
6. change in pressure:
   [latex]\begin{array}{cccc}\text{Ni}\left(s\right)+\hfill & 4\text{CO}\left(g\right)\hfill & \rightleftharpoons\hfill & \text{Ni}{\left(\text{CO}\right)}_{4}\left(g\right)\hfill \\ & 4x\hfill & & x\hfill \\ & \hfill 0.40\text{atm}\hfill & & 0.1\text{ atm}\hfill \end{array}[/latex]

3. Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.

1. (a) Write the starting conditions, change, and equilibrium constant in tabular form.

Since K is very small, ignore x in comparison with 0.129 M. The equilibrium expression is

[latex]\begin{array}{l} {K}_{c}=1.07\times {10}^{-5}=\frac{{\left[{\text{NO}}_{2}\right]}^{2}}{\left[{\text{N}}_{2}{\text{O}}_{4}\right]}=\frac{{\left(2x\right)}^{2}}{\left(0.129-x\right)}=\frac{{\left(2x\right)}^{2}}{0.129}\hfill \\ {x}^{2}=\frac{0.129\times 1.07\times {10}^{-5}}{4}=3.45\times {10}^{-7}\hfill \end{array}[/latex]

x = 5.87 × 10⁻⁴

The concentrations are:

• [NO₂] = 2x = 5.87 × 10^-4 = 1.17 × 10^-3M
• [N₂O₄] = 0.129 – x = 0.129 – 5.87 × 10^-4 = 0.128 M

(b) Percent error [latex]=\frac{5.87\times {10}^{-4}}{0.129}\times 100\%=0.455\%[/latex]. The change in concentration of N₂O₄ is far less than the 5% maximum allowed.

3.  (a) Write the balanced equation, and then set up a table with initial pressures and the changed pressures using x as the change in pressure. The simplest way to find the coefficients for the x values is to use the coefficient in the balanced equation.

[latex]{K}_{P}=2.2\times {10}^{-6}=\frac{\left({P}_{{\text{S}}_{2}}\right){\left({P}_{{\text{H}}_{2}}\right)}^{2}}{{\left({P}_{{\text{H}}_{2}\text{S}}\right)}^{2}}=\frac{\left[x\right]{\left[2x\right]}^{2}}{{\left[0.824-2x\right]}^{2}}[/latex]

Since the value of K_p is much smaller than 0.824, the 2x term in 0.824 − 2x is deemed negligible and, therefore, can be dropped.

[latex]\begin{array}{l}{ }2.2\times {10}^{-6}=\frac{\left[x\right]{\left[2x\right]}^{2}}{{\left[0.824\right]}^{2}}\hfill \\ 2.2\times {10}^{-6}=\frac{4{\text{X}}^{3}}{\left[0.679\right]}\hfill \end{array}[/latex]

• 1.494 × 10⁻⁶ = 4x³
• 3.73 × 10⁻⁷ = x³
• 7.20 × 10⁻³ = x

Final equilibrium pressures:

• [H₂S] = 0.824 − 2x = 0.824 − 2(7.20 × 10⁻³) = 0.824 – 0.0144 = 0.810 atm
• [H₂] = 2x = 2(7.2 × 10⁻³) = 0.014 atm
• [S₂] = [x] = 0.0072 atm

(b) The 2x is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2x is [latex]\frac{0.014}{0.824}\times 100\%=1.7\%[/latex], which is less than allowed by the “5% test.” The error is, indeed, negligible.

5. As all species are gases and are in M concentration units, a simple K_c equilibrium can be solved using the balanced equation:

[latex]{K}_{c}=0.0211=\frac{\left[{\text{PCl}}_{3}\right]\left[{\text{Cl}}_{2}\right]}{\left[{\text{PCl}}_{5}\right]}=\frac{\left[x\right]\left[x\right]}{\left[2.00-x\right]}[/latex]

As the value of K_c is substantial when compared with 2.00 M, the x terms in 2.00 − x cannot be disregarded. Thus, x must be solved by using the quadratic formula.

[latex]0.0211=\frac{{x}^{2}}{\left[2.00-x\right]}=0.0422-0.0211x={x}^{2}[/latex]

Begin by arranging the terms in the form of the quadratic equation:

• ax² + bx + c = 0
• x² + 0.0211x − 0.0422 = 0

Next, solve for x using the quadratic formula.

[latex]x=\frac{\text{-}b\pm\sqrt{{b}^{2}-4ac}}{2a}=\frac{-0.0211\pm\sqrt{{\left(0.0211\right)}^{2}-4\left(1\right)\left(-0.0422\right)}}{2\left(1\right)}[/latex]

[latex]=\frac{-0.0211\pm\sqrt{0.0004452+0.1688}}{2}=\frac{-0.0211\pm 0.4114}{2}[/latex]

= 0.195 M or −0.216 M

The process of dissociation renders only positive quantities. Thus, x must be a positive value when factored into the solution so as to guarantee a realistic result. The final equilibrium concentrations are: [PCl₃] = 2.00 − x = 2.00 − 0.195 = 1.80 M; [PC₃] = [Cl₂] = x = 0.195 M; [PCl₃] = [Cl₂] = x = 0.195 M.

2. Write the equilibrium constant expression and solve for P_BrCl.

[latex]\begin{array}{l}{K}_{P}=\frac{{\left({P}_{\text{BrCl}}\right)}^{2}}{{P}_{{\text{Cl}}_{2}}{P}_{{\text{Br}}_{2}}}=\frac{{\left({P}_{\text{BrCl}}\right)}^{2}}{\left(0.115\right)\left(0.450\right)}=4.7\times {10}^{-2}\\ {\left({P}_{\text{BrCl}}\right)}^{2}=0.115\times 0.450\times 4.7\times {10}^{-2}=2.43\times {10}^{-3}\text{atm}\end{array}[/latex]

P_BrCl = 4.9 × 10^-2 atm

4. Because two of the substances involved in the equilibrium are solids, their activities are 1, and their pressures are constant and do not appear in the equilibrium expression.

[latex]\begin{array}{l}{K}_{p}=4.08\times {10}^{-25}={\left({P}_{{\text{H}}_{2}\text{O}}\right)}^{10}\\ {P}_{{\text{H}}_{2}\text{O}}=\sqrt[10]{4.08\times {10}^{-25}}=3.64\times {10}^{-3}\text{atm}\end{array}[/latex]

7. As the equilibrium constant for the reaction is large at ≈ 10⁵, first assume that the reaction is complete—that is, that one or both reactants are completely consumed.

As only NO₂ exists at this stage, the reaction will establish equilibrium:

As this equilibrium is the reverse of that originally given, the new K_c will be the inverse of the given K_c—that is, [latex]\frac{1}{{K}_{c}}[/latex].

[latex]\begin{array}{ll}{K}_{\text{new}}\hfill & =\frac{1}{{K}_{c}}=\frac{{\left[\text{NO}\right]}^{2}\left[{\text{O}}_{2}\right]}{{\left[{\text{NO}}_{2}\right]}^{2}}\hfill \\ \hfill & =\frac{1}{2.3\times {10}^{5}}=4.35\times {10}^{-6}=\frac{{\left[2x\right]}^{2}\left[x\right]}{{\left[0.20-2x\right]}^{2}}\hfill \end{array}[/latex]

Because the new equilibrium constant is small compared with the 0.2 M term, the 2x can be dropped and the equilibrium solved.

[latex]4.35\times {10}^{-6}=\frac{\left(4{x}^{2}\right)x}{\left[0.04\right]}[/latex]

4x³ = 1.74 × 10⁻⁷

x³ = 4.35 × 10⁻⁸

x = 3.52 × 10⁻³

Final equilibrium concentrations:

• [NO₂] = 0.20 − 2x = 0.20 − 2(0.00352) = 0.19 M
• [NO] = 2x = 2(0.00352) = 0.0070 M
• [O₂] = x = 0.0035 M

2x is small compared to 0.20 M because

[latex]\left(\frac{2x}{0.20}\right)\times 100\%=\left(\frac{0.00352}{0.20}\right)\times 100\%=1.7\%[/latex] which is less than the maximum 5% allowed.

9. Assume that the reaction goes to completion, where one or both reactants are completely used up:

Now assume the equilibrium will be established by the reaction moving toward the reactants:

[latex]{K}_{P}=\frac{{P}_{{\text{NO}}_{2}}{P}_{{\text{O}}_{2}}}{{P}_{{\text{O}}_{3}}{P}_{\text{NO}}}=\frac{\left(1.2\times {10}^{-8}-x\right)\left(1.2\times {10}^{-8}-x\right)}{\left(x\right)\left(x\right)}[/latex]

As K_P is much larger than 1.2 × 10⁻⁸, the X terms in the expression 1.2 × 10⁻⁸ − x are relatively negligible and, therefore, can be dropped.

[latex]\begin{array}{l}{ }6.0\times {10}^{34}=\frac{\left(1.2\times {10}^{-8}\right)\left(1.2\times {10}^{-8}\right)}{\left(x\right)\left(x\right)}\hfill \\ {x}^{2}=\frac{\left(1.44\times {10}^{-16}\right)}{6.0\times {10}^{34}}=2.4\times {10}^{-51}\hfill \end{array}[/latex]

x = 4.9 × 10⁻²⁶ atm

[latex]{P}_{{\text{O}}_{3}}=x=4.9\times {10}^{-26}\text{atm}[/latex]

Clearly x is much smaller than 1.2 × 10⁻⁴ so the assumption is valid.

11. The number of moles of I₂ is

[latex]\begin{array}{l}{ }\text{mol}=\frac{63.5\text{g}}{253.809\text{g}{\text{mol}}^{-1}}=0.250\text{mol}{\text{I}}_{2}\hfill \\ {K}_{c}=\frac{{\left[\text{HI}\right]}^{2}}{\left[{\text{H}}_{2}\right]\left[{\text{I}}_{2}\right]}\hfill \end{array}[/latex]

The unit for each concentration term is moles per liter. If the volume were known for this exercise, the number of moles in each term should be divided by this volume. However, there are two terms in the numerator and two terms in the denominator, so these volumes cancel one another. Consequently, for any expression with the same number of numerator terms as denominator terms, the number for moles can be used in place of moles per liter. In this exercise, the volume is not needed even though it is given.

(mol HI)² = K × mol H₂ × mol I₂

= 50.2 × 1.25 mol × 0.250 mol

= 15.7 mol²

mol HI = [latex]\sqrt{15.7{\text{mol}}^{2}}=3.96\text{mol}[/latex]

Mass(HI) = 3.96 mol × 127.9124 g/mol = 507 g

13. At equilibrium the concentration of CO₂ = K_c = 0.50 M. The number of moles CO₂ in the system is then mol CO₂ = 6.5 L × 0.50 mol/L = 3.3 mol. The minimum moles of CaCO₂ required is then just more than:

[latex]3.3\text{mol}{\text{CO}}_{2}\times \frac{1\text{mol}{\text{CaCO}}_{3}}{1\text{mol}{\text{CO}}_{2}}=3.3\text{mol}{\text{CaCO}}_{3}[/latex]

[latex]3.3\text{mol}{\text{CO}}_{2}\times \frac{1\text{mol}{\text{CaCO}}_{3}}{100.1\text{g}{\text{CaCO}}_{3}}=330\text{g}[/latex]

1. (a) The reaction is [latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\text{.}[/latex] At equilibrium, [latex]{K}_{P}=\frac{\left({P}_{{\text{N}}_{2}{\text{O}}_{4}}\right)}{{\left({P}_{{\text{NO}}_{2}}\right)}^{2}}\text{.}[/latex] The value of K_P must remain the same when the pressure increases to 9.0 atm. Both gases must increase in pressure.

(b) [latex]2{\text{NO}}_{2}\left(g\right)\rightleftharpoons{\text{N}}_{2}{\text{O}}_{4}\left(g\right)\text{.}[/latex] At equilibrium, [latex]{K}_{P}=\frac{{P}_{{\text{N}}_{2}{\text{O}}_{4}}}{{\left({P}_{{\text{NO}}_{2}}\right)}^{2}}=\frac{0.70}{{\left(0.30\right)}^{2}}=\frac{0.70}{0.09}=7.78.[/latex] For a total pressure of 9.0 atm, the pressure of N₂O₄ is 9.0 – x; that of NO is x.

[latex]{K}_{P}=\frac{9.0-x}{{x}^{2}}=7.78[/latex]

7.78x² + x − 9.0 = 0

Use the quadratic expression, where

[latex]x=\frac{\text{-}b\pm \sqrt{{b}^{2}-4ac}}{2a}=\frac{-1\pm \sqrt{1+4\left(7.78\right)\left(-9.0\right)}}{2\left(7.78\right)}.[/latex]

[latex]=\frac{-1\pm 16.765}{15.56}=1.013[/latex]

For the plus sign (the negative sign gives a negative pressure which is impossible)

[latex]x=\frac{15.765}{15.56}=1.013[/latex]

The other answer is extraneous. The pressures are [latex]{P}_{{\text{N}}_{2}{\text{O}}_{4}}[/latex] = 8.0 atm and [latex]{P}_{{\text{NO}}_{2}}[/latex] = 1.0 atm

3. (a) For the above reaction, [latex]{K}_{c}=\frac{\left[{\text{CO}}_{2}\right]\left[{\text{H}}_{2}\right]}{\left[\text{CO}\right]\left[{\text{H}}_{2}\text{O}\right]}=5.0.[/latex] The concentrations at equilibrium are 0.20 M CO, 0.30 M H₂O, and 0.90 M H₂. Substitution gives [latex]K=5.0=\frac{\left[{\text{CO}}_{2}\right]\left[0.90\right]}{\left[0.20\right]\left[0.30\right]};[/latex] [latex]\left[{\text{CO}}_{2}\right]=\frac{5.0\left(0.20\right)\left(0.30\right)}{0.90}=0.33M;[/latex] Amount of CO₂ = 0.33 mol × 1 = 0.33 mol.

(b) At the particular temperature of reaction, K_c remains constant at 5.0. The new concentrations are 0.40 M CO, 0.30 M H₂O, and 1.2 M H₂.

[latex]\frac{\left[{\text{CO}}_{2}\right]\left(1.2\right)}{\left(0.40\right)\left(0.30\right)}=5.0[/latex]

[latex]{\left[\text{CO}\right]}^{2}=0.50M[/latex]

Amount of CO₂ = 0.50 mol × 1 = 0.50 mol. Added H₂ forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H₂ is added.

5. P_H2 = 8.64 × 10⁻¹¹atm

P_O2 = 0.250atm

P_H2O = 0.500 atm

7. (a) [latex]{K}_{c}=\frac{{\left[{\text{NH}}_{3}\right]}^{4}{\left[{\text{O}}_{2}\right]}^{7}}{{\left[{\text{NO}}_{2}\right]}^{4}{\left[{\text{H}}_{2}\text{O}\right]}^{6}}\text{.}[/latex] (b) Because [NH₃] is in the numerator of K_c, [NH₃] must increase for Q_c to reach K_c. (c) That decrease in pressure would decrease [NO₂] because the NO₂ must convert to NH₃ to reduce the effects of the expansion and the consequent relative decrease in products. (d) The relative pressures are controlled by the stoichiometry of the reaction.

[latex]{P}_{{\text{O}}_{2}}=\frac{7}{4}{P}_{{\text{NO}}_{2}}=\frac{7}{4}\left(28\text{torr}\right)=49\text{torr}[/latex]

9. [fructose] = 0.15 M

11. Write the balanced equilibrium expression. With all of the species as gases, it is a straightforward K_P problem to solve. However, all species must be converted to pressures, from other units related to concentration. For N₂O₃, with 0.236 mol in 1.52 L at 25 ºC: PV = nRT

[latex]\begin{array}{lll}P& =& \frac{n}{V}RT\\ & =& \frac{0.236\cancel{\text{mol}}}{1.52\cancel{\text{L}}}\times \frac{\left(0.08206\cancel{\text{L}}\text{atm}\right)\left(298.15\cancel{\text{K}}\right)}{\cancel{\text{mol}}\cancel{\text{K}}}\\ & =& 3.80\text{atm}\end{array}[/latex]

Write the balanced equation and the equilibrium changes:

[latex]{K}_{P}=\frac{\left({P}_{\text{NO}}\right)\left({P}_{{\text{NO}}_{2}}\right)}{\left({P}_{{\text{N}}_{2}{\text{O}}_{3}}\right)}[/latex]

[latex]1.91\text{atm}=\frac{\left(x\right)\left(x\right)}{\left(3.80-x\right)}[/latex]

Because 3.80 is substantial when compared with the equilibrium constant, the value of X must be considered. A quadratic equation must be solved:

[latex]1.91=\frac{{x}^{2}}{\left(3.80-x\right)}[/latex]

7.258 − 1.91x = x²

0 = x² +1.91x – 7.258

0 = ax² + bx + c

[latex]\begin{array}{lll}x& =& \frac{\text{-}b\pm \sqrt{{b}^{2}-4ac}}{2a}\\ & =& \frac{-1.91\pm \sqrt{{\left(1.91\right)}^{2}-4\left(1\right)\left(-7.258\right)}}{2\left(1\right)}\\ & =& \frac{-1.91\pm \sqrt{3.648+29.032}}{2}\\ & =& \frac{-1.91\pm \sqrt{32.680}}{2}\\ & =& \frac{-1.91\pm 5.717}{2}\end{array}[/latex]

= 1.90 atm or -3.81 atm

As negative pressure is not possible in this case, we use the positive value only. The final pressures are: [latex]{P}_{{\text{N}}_{2}{\text{O}}_{3}}[/latex] = 3.80 − x = 3.80 − 1.90 = 1.90 atm and [latex]{P}_{\text{NO}}={P}_{{\text{NO}}_{2}}={P}_{{\text{NO}}_{2}}x=1.90\text{atm}[/latex]

13.  (a) Recall that osmotic pressure is a colligative property, depending on the total number of particles present in the system. The osmotic pressure equation is π = iMRT. Given that the two acids face identical reaction conditions, the osmotic pressure (π) will increase only with a corresponding (>1) in the degree of ionization (i). Because both acids produce two ions per molecule dissociated, the number of particles in solution depends on the size of their respective equilibrium constants (K_c(HA) and K_c(HB)):

[latex]\text{HA}\rightleftharpoons{\text{H}}^{+}+{\text{A}}^{-}\text{HB}\rightleftharpoons{\text{H}}^{+}+{\text{B}}^{-}[/latex]

Because HB has a greater osmotic pressure (0.345 atm) than HA (0.293 atm), HB must produce more particles. Therefore, HB ionizes to a greater degree and has the larger K_c.

(b) Determine the value of i for each acid. For HA,

[latex]\pi =iMRT=0.293\text{atm}=i\left(0.010M\right)\left(\frac{0.08206\text{L atm}}{\text{mol K}}\right)\left(298.15\text{K}\right)[/latex]

i_HY = 1.2

For HB,

[latex]\pi =iMRT=0.345\text{atm}=i\left(0.010M\right)\left(\frac{0.08206\text{L atm}}{\text{mol K}}\right)\left(298.15\text{K}\right)[/latex]

i_HZ = 1.4

The i values mean that for every mole of originally undissociated species, i moles particles are produced in solution. The dissociations can be expressed in tabular form. For HA,

0.012 M = xM + xM + (0.010 − x) M

x = 0.002 M

For HB, the only difference from the table for HA is the value of i_HB. x = 0.004 M.

Now calculate the values of K_c(HA) and K_c(HB).

For HA,

[latex]{K}_{c}\left(\text{HA}\right)=\frac{\left[x\right]\left[x\right]}{\left[0.010-x\right]}=\frac{\left(0.002\right)\left(0.002\right)}{0.008}=5\times {10}^{-4}[/latex]

For HB,

[latex]{K}_{c}\left(\text{HB}\right)=\frac{\left[x\right]\left[x\right]}{\left[0.010-x\right]}=\frac{\left(0.004\right)\left(0.004\right)}{0.006}=3\times {10}^{-3}[/latex]