Butyric acid, \(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\), is responsible for the odor of rancid butter and cheese. Its \(K_{\mathrm{a}}\) is \(1.51 \times 10^{-5} .\) Calculate \(\left[\mathrm{H}^{+}\right]\) in solutions prepared by adding enough water to the following to make \(1.30 \mathrm{~L}\). (a) \(0.279 \mathrm{~mol}\) (b) \(13.5 \mathrm{~g}\)

The information given for part (a) of this exercise is: - Moles of butyric acid (\(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\)): 0.279 mol - Volume of solution: 1.3 L - \(K_\mathrm{a}\) of butyric acid: \(1.51 \times 10^{-5}\) First, we need to calculate the initial concentration of butyric acid in the solution: $$[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]_0 = \frac{0.279 \ \text{mol}}{1.3 \ \text{L}}$$

Divide the moles of butyric acid by the volume of the solution to find the initial concentration: $$[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]_0 = \frac{0.279 \ \text{mol}}{1.3 \ \text{L}} = 0.2146 \ \mathrm{M}$$

We can write the reaction of butyric acid with water to release \(\mathrm{H}^+\) ions as follows: \(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}(aq) \leftrightharpoons \mathrm{H}^+(aq) + \mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^-(aq)\) Now, we can set up an ICE (Initial, Change, Equilibrium) table to help us keep track of the concentrations of each species in the solution: ||||| |---|---|---|---| ||Initial (M)|Change (M)|Equilibrium (M)| |\(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\)|0.2146|-x|0.2146 - x| |\(\mathrm{H}^+\)|0|+x|x| |\(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^-\)|0|+x|x| Now, we can use the given \(K_\mathrm{a}\) value and the equilibrium expression for the reaction: $$K_\mathrm{a} = \frac{[\mathrm{H}^+][\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]}$$ Substitute the equilibrium concentrations from the ICE table into the expression: $$1.51 \times 10^{-5} = \frac{x\cdot x}{0.2146-x}$$

Since the \(K_\mathrm{a}\) value is small, we can assume that the dissociation of butyric acid is minimal. Hence, we can simplify the expression by assuming that the subtraction of x in the denominator doesn't significantly change the value: $$1.51 \times 10^{-5} \approx \frac{x^2}{0.2146}$$ Now, solve for x (concentration of \(\mathrm{H}^+\) ions): $$x^2 = (1.51 \times 10^{-5})(0.2146)$$ $$x = \sqrt{(1.51 \times 10^{-5})(0.2146)}$$ $$x = 9.77 \times 10^{-4} \ \mathrm{M}$$ So, the concentration of \(\mathrm{H}^+\) ions in the solution for part (a) is \(9.77 \times 10^{-4} \ \mathrm{M}\).

For part (b), we are given 13.5 grams of butyric acid instead of moles. We will first convert grams to moles: Molar mass of butyric acid: $$\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2} = 12.01 + 4(12.01) + 7(1.01) + 2(16.00) = 88.11 \ \text{g/mol}$$ $$\text{moles of butyric acid} = \frac{13.5 \ \text{g}}{88.11 \ \text{g/mol}}$$

Divide the grams of butyric acid by its molar mass to find the moles: $$\text{moles of butyric acid} = \frac{13.5 \ \text{g}}{88.11 \ \text{g/mol}} = 0.1532 \ \mathrm{mol}$$ Now, we can repeat the same calculation process from part (a), but using 0.1532 moles of butyric acid instead of 0.279 moles.

Following the same steps as in part (a), we calculate the initial concentration, set up the ICE table, use the equilibrium expression, and solve for the concentration of \(\mathrm{H}^+\) ions: Initial concentration of butyric acid: $$[\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}]_0 = \frac{0.1532 \ \text{mol}}{1.3 \ \text{L}} = 0.1178 \ \mathrm{M}$$ The equilibrium expression becomes: $$K_\mathrm{a} = \frac{x^2}{0.1178-x}$$ By simplifying the expression and solving for x (concentration of \(\mathrm{H}^+\) ions), we obtain: $$x^2 = (1.51 \times 10^{-5})(0.1178)$$ $$x = \sqrt{(1.51 \times 10^{-5})(0.1178)}$$ $$x = 7.45 \times 10^{-4} \ \mathrm{M}$$ So, the concentration of \(\mathrm{H}^+\) ions in the solution for part (b) is \(7.45 \times 10^{-4} \ \mathrm{M}\).