Question

$ 100{\text{ }}ml $ of $ 0.01{\text{ }}M $ solution of $ NaOH $ is diluted to $ 1d{m^3} $ . What is the $ pH $ of the diluted solution?
(A) $ 12 $
(B) $ 11 $
(C) $ 2 $
(D) $ 3 $

Answer 1

Hint :Concentration of the solution depends upon temperature and volume of the solution. It changes with change in volume i.e. concentration is inversely proportional to volume of the solution.

Complete Step By Step Answer:
Given: $ 100ml, $ $ 0.01{\text{ }}M{\text{ }}NaOH $ solution is diluted to $ 1d{m^3} $
i.e. $ 10 $ times diluted $ NaOH $ solution.
 $ \therefore $ The resultant concentration of the NaOH is given by
 $ NaOH = \dfrac{{0.01M}}{{10}} = 0.001M $
 $
  NaOH\xrightarrow{{}}{\text{ }}N\mathop a\limits^ + {\text{ }} + {\text{ }}O\mathop H\limits^ - \\
  {\text{0}}{\text{.001M 0 0 }} \\
  {\text{ 0 0}}{\text{.001M 0}}{\text{.001M}} \\
  $
We know that we have
 $
  \therefore conc.{\text{ of OH}} = 0.001M \\
  pOH = - \log \left[ {O\mathop H\limits^ - } \right] \\
  $
By using the formula, we get
 $
  \therefore pOH = - \log \left( {0.001} \right) \\
   \Rightarrow pOH = - \log \left( {{{10}^{ - 3}}} \right) \\
  $
On using the property of log we get
 $
  pOH = - \left( { - 3} \right){\text{ log 10}} \\
  pOH = 3{\text{ }}\left( {\because {\text{ log }}10 = 1} \right) \\
  $
By using the formula,
 $ pH + pOH = 14 $
 $
  pH = 14 - pOH \\
  pH = 14 - 3 \\
  $
On further solving we get
 $ pH = 11 $
Hence, $ pH $ of the resulting solution $ = 11 $
Therefore the $ pH $ of the diluted solution is 11.
Then, the correct option is B.

Note :
We have studied that the nature of a solution whether neutral, acidic or basic is expressed in terms of $ \left[ {{H_3}{O^ + }} \right] $ by scientific notation as $ 1.0 \times {10^{ - x}} $ . Here, $ x $ may vary from $ 0 $ to $ 14 $ at $ 298{\text{ }}K $ . But the way of expressing concentration is quite inconvenient. For example, if the $ \left[ {{H_3}{O^ + }} \right] $ of a solution is $ 3.5 \times {10^{ - 4}}M $ , this means that actually it is $ 0.00035M $ .
In order to simplify it, Sorensen in $ 1909 $ has suggested another method for expressing the $ \left[ {{H_3}{O^ + }} \right] $ . It is called $ pH $ and the scale of measurement as pH scale. The $ pH $ of a solution is defined as the negative logarithm of its hydronium ion concentration.