Acid-base reactions usually go to completion. Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) a strong acid reacting with a strong base, (b) a strong acid reacting with a weak base \(\left(\mathrm{NH}_{3}\right),\) (c) a weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a strong base, \((\mathrm{d})\) a weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) reacting with a weak base \(\left(\mathrm{NH}_{3}\right)\) (Hint: Strong acids exist as \(\mathrm{H}^{+}\) ions and strong bases exist as \(\mathrm{OH}^{-}\) ions in solution. You need to look up the \(K_{\mathrm{a}}, K_{\mathrm{b}}\), and \(K_{\mathrm{w}}\) values.)

Step by step solution

01

(a) Strong Acid with Strong Base

The reaction for a strong acid with a strong base goes to completion. Thus, the reaction is: \(HA + BOH \rightarrow H_2O + BA\). The equilibrium constant for this reaction is \(K\), but since the reaction goes to completion, we don't need the precise value of \(K\).

02

(b) Strong Acid with Weak Base

The reaction of a strong acid with a weak base also goes to completion: \(HA + B \leftrightarrow BH^+ + A^- \). However, the base will also have an equilibrium with water: \(B + H_2O \leftrightarrow BH^+ + OH^-\). The equilibrium constants are \(K_a\) for the acid and \(K_b\) for the base. The overall \(K\) can be obtained by multiplying \(K_a\) and \(K_b\) and they should equal to \(K_w\). Therefore, these reactions also go to completion.

03

(c) Weak Acid with Strong Base

The reaction of a weak acid with a strong base is given by: \(HA + BOH \leftrightarrow A^- + H_2O\). The equilibrium constant for this reaction is \(K'\), and it involves the base ionization of water. However, as the base is a strong base and the reaction goes to completion, we don't need the precise value of \(K'\).

04

(d) Weak Acid with Weak Base

The reaction of a weak acid with a weak base is given by: \(HA + B \leftrightarrow BH^+ + A^-\). In this case, both the acid and the base are weak. Therefore, both will have their own equilibrium constants \(K_a\) and \(K_b\). The overall \(K\) can be obtained by multiplying \(K_a\) and \(K_b\), and they should equal to \(K_w\). Therefore, in the end these reactions go to completion as well.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!