Question

What is the $ pH $ of the $ NaOH $ solution when $ 0.04{\text{ gm }} $ of it dissolved in water and made to $ 100{\text{ ml }} $ solution $ ? $
 $ (i){\text{ 2}} $
 $ (ii){\text{ 1}} $
 $ (iii){\text{ 13}} $
 $ (iv){\text{ 12}} $

Answer 1

Hint: Here we are given a mass of $ NaOH $ in solution. We find its molarity using the volume of solution. Then we will find out the concentration of $ \left[ {O{H^{ - 1}}} \right] $ ions. Then we will find the value of $ pH $ using the relation between $ pH $ and $ pOH $ .
 $ (i){\text{ pOH = - log}}\left[ {O{H^ - }} \right] $
 $ (ii){\text{ pH + pOH = 14}} $ .

Complete answer:
We have to find the $ pH $ of the solution of $ NaOH $ . Since we know that $ NaOH $ is basic in nature thus $ pH $ cannot be calculated directly using the formula. Thus by following the certain steps, $ pH $ of the given solution can be calculated.
Step $ (1) $ : Calculation of $ \left[ {O{H^{ - 1}}} \right] $ ions.
We will find the molarity of the $ NaOH $ solution as,
Given mass of $ NaOH{\text{ = 0}}{\text{.04 gm}} $
Molar mass of $ NaOH{\text{ = }}\left( {{\text{23 + 16 + 1}}} \right){\text{ gm}} $
Molar mass of $ NaOH{\text{ = 40 gm}} $
Volume of solution $ {\text{ = 100 ml = 0}}{\text{.1 L}} $
Molarity $ = {\text{ }}\dfrac{{given{\text{ mass}}}}{{molar{\text{ mass}}}}{\text{ }} \times {\text{ }}\dfrac{1}{{volume{\text{ of solution in Litre}}}} $
On substituting the values we get,
Molarity $ = {\text{ }}\dfrac{{0.04}}{{40}}{\text{ }} \times {\text{ }}\dfrac{1}{{0.1}}{\text{ M}} $
Molarity $ = {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}} $
Since it is mono-basic compound thus we can say that the concentration of $ \left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 2}} $ .
Step $ (ii) $ Calculation of $ pH $
We get the $ \left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 2}} $ . Now we can find the $ pOH $ , then we can find $ pH $ using the relation. Therefore, we know that,
 $ {\text{ pOH = - log}}\left[ {O{H^ - }} \right] $
 $ {\text{ pOH = - log}}\left( {{{10}^{ - 2}}} \right) $
 $ {\text{ pOH = 2 log10}} $
 $ {\text{ pOH = 2}} $
Since we know that, for any solution $ {\text{ pH + pOH = 14}} $ . Thus using this relation we can find $ pH $ as,
 $ {\text{ pH = 14 - pOH}} $
 $ pH{\text{ = 14 - 2}} $
 $ pH{\text{ = 12}} $
Thus $ pH $ of the given $ NaOH $ solution is $ 12 $ . Thus the correct option is $ (iv){\text{ 12}} $ .

Note:
$ NaOH $ solution is basic in nature. Therefore its $ pH $ will be greater than seven. If it comes to less than seven then check the calculations again. Since $ NaOH $ is a mono-basic compound which means it gives only one mole of hydroxide ions. Thus the molarity of $ NaOH $ is equivalent to concentration of its constituent ions. Here we use the base of the log as $ 10. $