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Question: What is the pH of a 1M \( C{{H}_{3}}COOH \) solution? ( \( {{K}_{a}} \) of acetic acid \( =1.8\times...

What is the pH of a 1M CH3COOHC{{H}_{3}}COOH solution? ( Ka{{K}_{a}} of acetic acid =1.8×10−5,Kw=10−14mol2litre−2=1.8\times {{10}^{-5}},{{K}_{w}}={{10}^{-14}}mo{{l}^{2}}litr{{e}^{-2}} )

A. 9.49.4

B. 4.84.8

C. 3.63.6

D. 2.42.4

Explanation

Solution

pHpH is the term where we measure acidity or basicity of a liquid solution, the values of the concentration of hydrogen ion varies between ranges 1−10−141-{{10}^{-14}} gram equivalents per litre into numbers between 0−140-14 .

If a solution has a lower value than seven, it will be considered an acidic solution, similarly, if a solution has a higher value than seven, it will be considered a basic solution, and if it has exactly seven, then it will be a neutral solution.

Complete step by step answer:

In order to calculate the pHpH of a 1M1M CH3COOHC{{H}_{3}}COOH solution, first, we need to look at the equilibrium which is being established in the following question. It can be chemically expressed as,

CH3COOH⇌CH3COO−+H+C{{H}_{3}}COOH\rightleftharpoons CH3CO{{O}^{-}}+{{H}^{+}}

Here we can see that one mole of acetic acid dissociates into its corresponding ions which are acetate and hydrogen ions. Now if we calculate the concentration of this liberated hydrogen, we will be able to know the pHpH of the solution from that value.

According to Ostwald’s dilution law equation, for weak acid, as we know acetic acid is a weak acid,

[H+]=(C×Ka)[{{H}^{+}}]=\sqrt{(C\times {{K}_{a}})}

Where [H+][{{H}^{+}}] is the concentration of hydrogen ion, CC is the molar concentration of the solution and Ka{{K}_{a}} is the ionisation constant of the acid.

So now we will put the values of concentration which is one molar given to us, and the Ka{{K}_{a}} which is =1.8×10−5=1.8\times {{10}^{-5}} also given to us,

[H+]=1×1.8×10−5=0.004243M[{{H}^{+}}]=\sqrt{1\times 1.8\times {{10}^{-5}}}=0.004243M

Since we got the value of concentration of hydrogen ions, we can easily calculate the pHpH of the solution, by taking negative log of this value,

pH=−log⁡[H+]=−log⁡(0.004243)pH=-\log [{{H}^{+}}]=-\log (0.004243)

=2.4=2.4

So, the correct answer is Option D.

Note: When a weak acid is dissolved in water it does not dissociate totally, while in the case of strong acids like hydrochloric acid, the dissociation of acid is complete, meaning, all the hydrogen atoms from the acid gets liberated from the acid in order to form hydronium ion.