Question

Atypicalaspirintabletcontains325mgofacetylsalicylic acid (HC9H7O4). Calculate the pH of a solution that is prepared by dissolving two aspirin tablets in one cup (237 mL) of solution. Assume the aspirin tablets are pure acetylsalicylic acid, Ka 5 3.3 3 1024.

Answer 1

`\boxed{2.65}`

Step-by-step explanation:

1. Mass of acetylsalicylic acid (ASA)

`m = \text{2 tablets} * \frac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}`

2. Moles of ASA

HC₉H₇O₄ =180.16 g/mol

`n = \text{750 mg} * \frac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}`

3. Concentration of ASA

`c = \frac{\text{4.163 mmol}}{\text{237 mL}} = \text{0.01757 mol/L}`

4. Set up an ICE table

`\begin{array}{ccccccc}\text{HA} & + & \text{H$_(2)$O}& \, \rightleftharpoons \, &\text{H$_(3)$O$^(+)$} & + &\text{A}^(-)\\0.01757 & & & &0 & & 0 \\-x & & & &+x & & +x \\0.01757-x & & & &x & & x \\\end{array}\\`

5. Solve for x

`K_{\text{a}} = \frac{\text{[H}_(3)\text{O}^(+)]\text{A}^(-)]} {\text{[HA]}} = 3.33 * 10^(-4)\\\\(x^(2))/(0.01757 - x) = 3.33 * 10^(-4)\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\( 0.01757 )/(3.33 * 10^(-4)) = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^(2) = 3.33 * 10^(-4)(0.01757 - x) \\\\x^(2) = 5.851 * 10^(-6) - 3.33 * 10^(-4)x\\\\x^(2) + 3.33 * 10^(-4)x - 5.851 * 10^(-6) = 0`

6. Solve the quadratic equation.

`a = 1; b = 3.33 * 10^(-4); c = -5.851 * 10^(-6)`

`x = (-b\pm√(b^2-4ac))/(2a)\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258`

7. Calculate the pH

`\rm [H_(3)O^(+)]= x \, mol \cdot L^(-1) = 0.002258 \, mol \cdot L^(-1)\\\text{pH} = -\log{\rm[H_(3)O^(+)]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}`