Question

calculate the$pH$, $pOH$ and $\left[ {{H^ + }} \right]ion$ concentration of
$0.015{\text{ }}m{\text{ }}HCl$.

Answer 1

[pH] is defined as the negative logarithm of ${H^ + }\;ion$ concentration, Hence the meaning of the name $pH$ is justified as the power of hydrogen as a$pH{\text{ }} = {\text{ }} - log{\text{ }}\left[ {{H^ + }} \right]$ . The concentration of the hydroxide ion can be expressed logarithmically by the $pOH$ . The $pOH$ of a solution is the negative logarithm of the hydroxide-ion concentration and represent as a $pOH{\text{ }} = {\text{ }} - log\left[ {OH{\;^{-\;}}} \right].$

Complete Step by step answer: Assuming that this is done under standard conditions
Hydrochloric acid is a strong acid, it dissociates into single $H + \;ions$ in an aqueous solution, which eventually become ${H_3}O + \;ions$ due to the presence of water $\left( {{H_2}O} \right).$
In this problem, We can assume that$\;HCl\;$ is completely ionized in water $pH$ is given by the equation,
$pH = - log{\text{ }}\left[ {{H^ + }} \right]$
$\left[ {{H^ + }} \right]\;$ is the hydrogen ion concentration in terms of Molarity.
$\;\left[ {{H^ + }} \right]\; = {\text{ }}0.015{\text{ }}m\; = {\text{ }}\left( {1.5{\text{ }} \times {{10}^{ - 2}}} \right)$
Putting values of $\left[ {H + } \right]\;$
$pH = - log{\text{ }}\left[ {{H^ + }} \right]$
$\Rightarrow pH = {\text{ }} - log{\text{ }}\left[ {1.5 \times {{10}^{ - 2}}} \right]$
$\Rightarrow pH = {\text{ }}2 - {\text{ }}log\frac{3}{2}$
$\Rightarrow pH = {\text{ }}2 + log{\text{ }}2-log{\text{ }}3$ $\therefore log3 = {\text{ }}0.477,log2 = 0.3010$
$\Rightarrow pH = 2 + {\text{ }}0.30-0.477$
$\Rightarrow pH = \;2.3 - 0.48$
$\Rightarrow pH\; = 1.88$
we can use the following relationship -
$pH\; + {\text{ }}pOH\; = 14{\text{ }}at{\text{ }}{25^ \circ }Celcius$
$1.88{\text{ }} + pOH{\text{ }} = 14$
$pOH\; = 12.12$
Now the value of $pH$, $pOH$ and $\left[ {{H^ + }} \right]ion$ concentration of $0.015{\text{ }}m{\text{ }}HCl$ are
$pH$ $= \;\;1.88$
$pOH$ $= \;\;12.12$
\left[ {{H^ + }} \right]ion$$$$ = {\text{ }}1.5{\text{ }} \times {10^{ - 2}}

Note: The $pH$ of a solution varies from$0{\text{ }}to{\text{ }}14$ . Solutions having a value of pH ranging $0{\text{ }}to{\text{ }}7$on$\;pH$ scale are termed as acidic and for the value of $pH$ ranging $7{\text{ }}to{\text{ }}14$ on $pH$scale are known as basic solutions when $pH$ equal to $7$ on $pH$scale are known as neutral solutions. Solutions having the value of $pH$ equal to $0$ are known to be strongly acidic solutions and solutions with the value of $pH$ equal to $14$ are termed as strongly basic solutions.