Question

The $pH$ of $0.001M \;C{H_3}COOH$ is:
(A) 3
(B) 11
(C) 3-7
(D) 7

Answer 1

Hint:
$pH$of solution defined as the magnitude of negative power to which 10 must be raised to compare the ${H^ + }$concentration.
Formula: $pH\;=\; - \log [{H^ + }]$

Complete Step by Step Solution:
$C{H_3}COOH$ dissociates as
$C{H_3}COOH + {H_2}O\underset {} \leftrightarrows C{H_3}CO{O^ - } + {H_3}{O^ + }$
$\therefore [{H^ + }] = [C{H_3}COOH]$
Since $[C{H_3}COOH] = 0.001$
Therefore, $[{H^ + }] = 0.001M$
$ = 1 \times {10^{ - 3}}M$

We know that
a. $\log 1 = 0$
b. $ \log 10 = 1$

$pH = - \log [{H^ + }]$
$\Rightarrow pH = - \log 1 \times {10^{ - 3}}$
$\Rightarrow pH = - \log 1 + 3\log 10$
$\therefore pH = 0 + 3 \times 1$
$pH = 3$

$\therefore pH$of $0.001M\;C{H_3}COOH $ solution is $3.$
$pH$ value of solution decides the acidic and basic nature of solution.
$pH$range is taken as $0$ to $14.$ The acidity or alkalinity represented as follows.

The $pH$ value of solution does not give the exact value of their relative strength.
Example:
If $pH$ of solution is \[1\], the solution has hydrogen ion concentration \[100\] times. Than that of a solution of $pH = 3$ and not \[3\] times.

Additional Information:
Acidic strength of acid increases when $pH$ of solution decreases.
Solution of $pH$$1$ is striges acid is an solution of $pH$$4.$
Solution of $pH$$7$ is neutral.
Solution of $pH$$9$ is less basic than solution of $pH$$12.$
Therefore, As $pH$ of solution increases form $8$ to $14$ alkalinity of solution increases.

Note:
Acetic acid is weak acid because it does not dissociate completely in aqueous solution.
It establishes equilibrium between insisted and unionized ions.
$C{H_3}COOH\underset {} \leftrightarrows C{H_3}CO{O^ - } + {H^ + }$
$pH$ is a measure of the relative amount of free hydrogen and hydroxyl ions in the water.