Question

Calculate\[\;pH\]value of \[0.0001N\;NaOH\;\] solution?

Answer 1

Hint: \[pH\] 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. \[pH = - log\left[ {{H^ + }} \right]\].