Question

Calculate the pH of $ 0.02 \cdot mol \cdot d{m^{ - 3}}$ of Calcium hydroxide?

Answer 1

Hint: The calcium hydroxide is a basic compound so its pH will be greater than the seven. The pH represents the acidity and basicity of the solution. On a pH scale, the solution having pH value of one is most acidic and solution having pH value of fourteen is most basic. For the given question, first we need to calculate pOH and then we can calculate the pH.

Complete answer:
We are given that the concentration of calcium hydroxide is $ 0.02 \cdot mol \cdot d{m^{ - 3}}$ . Now, assuming calcium hydroxide to be a strong base, the dissociation will be complete and that can be represented as $ Ca{(OH)_2} \rightleftharpoons C{a^{2 + }} + 2O{H^ - }$
Now, if complete dissociation occurs, then, one stoichiometric equivalent of hydroxide ions will dissociate in proportion to the concentration of the salt.

So, the $[O{H^ - }]$can be calculated as –
$ [O{H^ - }] = 0.02mol \cdot {L^{ - 1}} \times \left( {\dfrac{{2O{H^ - }}}{{Ca{{(OH)}_2}}}} \right) \approx 0.04M$
Now, using $[O{H^ - }]$ , pOH value can be calculated as –
$ \eqalign{
  & pOH = - \log [O{H^ - }] \cr
  & pOH = - \log (0.04M) \cr
  & pOH = 1.397 \cr} $
Now, the calculated value of pOH is $1.397$ . Using the formula $pH + pOH = 14$ , we can calculate the pH of the given solution of calcium hydroxide. The expected result should be greater than seven.
Now, we have $ pOH = 1.397$ , so using the formula, the pH will be given as –
$\eqalign{
  & pH = 14 - pOH \cr
  & pH = 14 - 1.397 \cr
  & pH = 12.603 \cr
  & pH \approx 12.6 \cr} $

Hence, the pH of $ 0.02 \cdot mol \cdot d{m^{ - 3}}$ of calcium hydroxide will be $12.6$ .

Note:
The pH calculation of bases is done by determining the dissociation and then calculating $[O{H^ - }]$ followed by calculating pOH. And, then using the formula $pH + pOH = 14$ to get the final value of pH. The value of pH for the strong bases will be near to fourteen.