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Question: \( 100{\text{ }}ml \) of \( 0.01{\text{ }}M \) solution of \( NaOH \) is diluted to \( 1d{m^3} \) . ...

100 ml100{\text{ }}ml of 0.01 M0.01{\text{ }}M solution of NaOHNaOH is diluted to 1dm31d{m^3} . What is the pHpH of the diluted solution?
(A) 1212
(B) 1111
(C) 22
(D) 33

Explanation

Solution

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,100ml, 0.01 M NaOH0.01{\text{ }}M{\text{ }}NaOH solution is diluted to 1dm31d{m^3}
i.e. 1010 times diluted NaOHNaOH solution.
∴\therefore The resultant concentration of the NaOH is given by
NaOH=0.01M10=0.001MNaOH = \dfrac{{0.01M}}{{10}} = 0.001M
NaOH→ Na+ + OH− 0.001M 0 0   0 0.001M 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
∴conc. of OH=0.001M pOH=−log⁡[OH−]  \therefore conc.{\text{ of OH}} = 0.001M \\\ pOH = - \log \left[ {O\mathop H\limits^ - } \right] \\\
By using the formula, we get
∴pOH=−log⁡(0.001) ⇒pOH=−log⁡(10−3)  \therefore pOH = - \log \left( {0.001} \right) \\\ \Rightarrow pOH = - \log \left( {{{10}^{ - 3}}} \right) \\\
On using the property of log we get
pOH=−(−3) log 10 pOH=3 (∵ log 10=1)  pOH = - \left( { - 3} \right){\text{ log 10}} \\\ pOH = 3{\text{ }}\left( {\because {\text{ log }}10 = 1} \right) \\\
By using the formula,
pH+pOH=14pH + pOH = 14
pH=14−pOH pH=14−3  pH = 14 - pOH \\\ pH = 14 - 3 \\\
On further solving we get
pH=11pH = 11
Hence, pHpH of the resulting solution =11= 11
Therefore the pHpH 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 [H3O+]\left[ {{H_3}{O^ + }} \right] by scientific notation as 1.0×10−x1.0 \times {10^{ - x}} . Here, xx may vary from 00 to 1414 at 298 K298{\text{ }}K . But the way of expressing concentration is quite inconvenient. For example, if the [H3O+]\left[ {{H_3}{O^ + }} \right] of a solution is 3.5×10−4M3.5 \times {10^{ - 4}}M , this means that actually it is 0.00035M0.00035M .
In order to simplify it, Sorensen in 19091909 has suggested another method for expressing the [H3O+]\left[ {{H_3}{O^ + }} \right] . It is called pHpH and the scale of measurement as pH scale. The pHpH of a solution is defined as the negative logarithm of its hydronium ion concentration.