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Question: What is the \( pH \) of the \( NaOH \) solution when \( 0.04{\text{ gm }} \) of it dissolved in wate...

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

Explanation

Solution

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

Complete answer:
We have to find the pHpH of the solution of NaOHNaOH . Since we know that NaOHNaOH is basic in nature thus pHpH cannot be calculated directly using the formula. Thus by following the certain steps, pHpH of the given solution can be calculated.
Step (1)(1) : Calculation of [OH−1]\left[ {O{H^{ - 1}}} \right] ions.
We will find the molarity of the NaOHNaOH solution as,
Given mass of NaOH = 0.04 gmNaOH{\text{ = 0}}{\text{.04 gm}}
Molar mass of NaOH = (23 + 16 + 1) gmNaOH{\text{ = }}\left( {{\text{23 + 16 + 1}}} \right){\text{ gm}}
Molar mass of NaOH = 40 gmNaOH{\text{ = 40 gm}}
Volume of solution  = 100 ml = 0.1 L{\text{ = 100 ml = 0}}{\text{.1 L}}
Molarity = given massmolar mass × 1volume of solution in Litre= {\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 = 0.0440 × 10.1 M= {\text{ }}\dfrac{{0.04}}{{40}}{\text{ }} \times {\text{ }}\dfrac{1}{{0.1}}{\text{ M}}
Molarity = 10−2 M= {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}}
Since it is mono-basic compound thus we can say that the concentration of [OH−1] = 10−2\left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 2}} .
Step (ii)(ii) Calculation of pHpH
We get the [OH−1] = 10−2\left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 2}} . Now we can find the pOHpOH , then we can find pHpH using the relation. Therefore, we know that,
 pOH = - log[OH−]{\text{ pOH = - log}}\left[ {O{H^ - }} \right]
 pOH = - log(10−2){\text{ pOH = - log}}\left( {{{10}^{ - 2}}} \right)
 pOH = 2 log10{\text{ pOH = 2 log10}}
 pOH = 2{\text{ pOH = 2}}
Since we know that, for any solution  pH + pOH = 14{\text{ pH + pOH = 14}} . Thus using this relation we can find pHpH as,
 pH = 14 - pOH{\text{ pH = 14 - pOH}}
pH = 14 - 2pH{\text{ = 14 - 2}}
pH = 12pH{\text{ = 12}}
Thus pHpH of the given NaOHNaOH solution is 1212 . Thus the correct option is (iv) 12(iv){\text{ 12}} .

Note:
NaOHNaOH solution is basic in nature. Therefore its pHpH will be greater than seven. If it comes to less than seven then check the calculations again. Since NaOHNaOH is a mono-basic compound which means it gives only one mole of hydroxide ions. Thus the molarity of NaOHNaOH is equivalent to concentration of its constituent ions. Here we use the base of the log as 10.10.