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Chapter 16: Problem 48

(a) Calculate the percent ionization of a \(0.20 \mathrm{M}\) solution of the monoprotic acetylsalicylic acid (aspirin). \(\left(K_{\mathrm{a}}=3.0 \times 10^{-4} .\right)(\mathrm{b})\) The \(\mathrm{pH}\) of gastric juice in the stomach of a certain individual is 1.00 . After a few aspirin tablets have been swallowed, the concentration of acetylsalicylic acid in the stomach is \(0.20 \mathrm{M}\) Calculate the percent ionization of the acid under these conditions.

Short Answer

Expert verified
The percent ionization of a 0.20 M solution of acetylsalicylic acid is 1.225%, while in the stomach it ionizes up to 50%.

Step by step solution

01

Setup the Equation

We can use the formula for the ionization of a weak acid : \(HA \leftrightarrow H^{+}+A^{-}\), Where, \(HA\) is the weak acid. We also know that the ionization constant \(Ka = \frac{[H^{+}][A^{-}]}{[HA]}\), where \([H^{+}]\) is the concentration of the H+ ion, \([A^-]\) is the concentration of the A- ion, and \([HA]\) is the initial concentration of the aciditylsalicyclic acid.
02

Calculation Percent ionization for a 0.20 M solution

Given that `Ka = 3.0 x 10^-4` and `HA = 0.20 M`. Setting up the ionization equation, `Ka = x^2/(0.20-x)`. Solving this results in x = 0.00245 M. Then, calculate the percent ionization using the formula \((\frac{x}{HA})\times 100%\), which results in \(1.225\%\).
03

Calculation for Gastric Juice

Knowing that pH = -log[H+] and the provided pH is 1.00, we can find that \([H+] = 10^{(-pH)}\), giving us \([H+] = 10^{-1}\) = 0.10. We can then calculate the percent ionization in the same way as in step 2, using \([H+]\) instead of x (as it is the same), and getting a percentage of \(50\%\).

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Most popular questions from this chapter

Explain why small, highly charged metal ions are able to undergo hydrolysis.

Describe this reaction according to the Lewis theory of acids and bases: $$\mathrm{AlCl}_{3}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AlCl}_{4}^{-}(a q)$$

Which of the following solutions has the highest \(\mathrm{pH} ?\) (a) \(0.40 \mathrm{M} \mathrm{HCOOH},\) (b) \(0.40 \mathrm{M} \mathrm{HClO}_{4}\) (c) \(0.40 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\).

A 0.0560 -g quantity of acetic acid is dissolved in enough water to make \(50.0 \mathrm{~mL}\) of solution. Calculate the concentrations of \(\mathrm{H}^{+}, \mathrm{CH}_{3} \mathrm{COO}^{-},\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) at equilibrium. \(\left(K_{\mathrm{a}}\right.\) for acetic acid \(=\) \(\left.1.8 \times 10^{-5} .\right)\)

The pOH of a solution is \(9.40 .\) Calculate the hydrogen ion concentration of the solution.
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