Question

Calculate$\;pH$value of $0.0001N\;NaOH\;$ solution?

Answer 1

$pH$ is a scale used to specify the acidity or basicity of an aqueous solution . It shows a negative logarithm of concentration of $H +$ ion. And is a$\;NaOH$ is a base act as a strong electrolyte (a strong electrolyte is a one which dissociates completely) so it completely dissociates into $N{a^ + }ion$ and $O{H^{ - {\text{ }}}}ion.$

Complete Step by step answer: Given : $0.0001{\text{ }}N{\text{ }}NaOH$
First we have to find its Molarity and the relation between the Normality and Molarity.
Molarity(M) refers to the concentration of a compound or ion in a solution and
Normality(N) refers to the molar concentration only of the acid component or only of the base component of the solution its formula is -
$Normality{\text{ }} = {\text{ }}molarity * n\;factor{\text{ }}$
($n{\text{ }}factor$ for the bases is equal to their acidity or the number of replaceable $O{H^ - }ion$ per molecule of the base )
so $n{\text{ }}factor$ of $NaOH{\text{ }} = {\text{ }}1\;$, then Normality is equal to Molarity so we are having $0.0001{\text{ }}M{\text{ }}NaOH$ which dissociates completely to give $$$$$$0.0001M{\text{ }}N{a^ + }{\text{ }}ion{\text{ }}and{\text{ }}0.0001{\text{ }}M{\text{ }}O{H^{ - {\text{ }}}}ion{\text{ }}.$We will use$O{H^ - }ion$concentration$\left( {0.0001} \right)$to calculate$pOH$(negative log of$O{H^ - }ion$concentration ) For$\begin{array}{*{20}{l}}
{pOH{\text{ }} = {\text{ }} - log\left[ {OH - } \right]} \\
{So,{\text{ }}with{\text{ }}a{\text{ }}\left[ {NaOH} \right]{\text{ }} = {\text{ }}0.0001,{\text{ }}the{\text{ }}pOH{\text{ }}is} \\
{pOH{\text{ }} = {\text{ }} - log\left( {0.0001} \right){\text{ }} = {\text{ }}4}
\end{array}$Normality is equal to Molarity.$\left[ {O{H^ - }} \right] = \left[ {NaOH} \right] = 0.0001N = 0.0001M$We get$pOH = - {\text{ }}log\left[ {O{H^ - }} \right] = - log\left[ {0.0001} \right] = 4$${\text{ }}\therefore {\text{pOH}} = - {\text{ l}}og\left( {{{10}^{ - 4}}} \right) = 4$We know that$\therefore pH{\text{ }} + {\text{ }}pOH{\text{ }} = 14$$so{\text{ }}pH{\text{ }} = 14-pOH{\text{ }}$$\Rightarrow pH = 14 - 4{\text{ }} = {\text{ }}10$Now the value of$pH{\text{ }}of{\text{ }}0.0001N.{\text{ }}NaOH$solution is$10$(alkaline) . sodium hydroxide is a strong base, it makes sense that the$pH$is above$;7$$ .

Note: The pH scale runs from$0{\text{ }}to{\text{ }}14$, a value of seven is considered neutral, less than seven acidic, and greater than seven basic. To calculate the value , take the log of a given hydrogen ion concentration and reverse the sign. $pH = - log\left[ {{H^ + }} \right]$.