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Question: Two solutions, A and B, each of \(100{\text{L}}\) was made by dissolving \(4{\text{g}}\) of \({\text...

Two solutions, A and B, each of 100L100{\text{L}} was made by dissolving 4g4{\text{g}} of NaOH{\text{NaOH}} and 9.8g9.8{\text{g}} of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} in water, respectively. The pH of the resultant solution obtained from mixing 40L40{\text{L}} of solution A and 10L10{\text{L}} of solution B is ______.

Explanation

Solution

To solve this question, you must use the concept of equivalents. Equal numbers of equivalents of an acid neutralize equal numbers of equivalents of a base. We shall calculate the molarity of the acid and base present and thus, the molarity of acid/base present after neutralization. The molarity of hydrogen ion or hydroxide ion present in the solution after neutralization shall be used to calculate the pH of the solution.
Formula used:
meq=M×V×x{{meq}} = {{M \times V \times }}x
Where meq{\text{meq}} is the number of milliequivalents.
M is the molarity of the acid / base solution
V is the volume of the solution
x is the n-factor of the acid or base

Complete step by step answer:
We know that, the acidity ofNaOH{\text{NaOH}} is equal to 1 while the basicity of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}is equal to 2, which are also the respective n factors of the two compounds.
The number of moles of a substance are equal to the ratio between its given mass and its molar mass.
The number of moles of NaOH{\text{NaOH}} are, n=440=0.1{\text{n}} = \dfrac{4}{{40}} = 0.1
The molarity is given as the number of moles per litre of the solution.
Molarity of NaOH{\text{NaOH}} solution =0.1100=10−3M = \dfrac{{0.1}}{{100}} = {10^{ - 3}}{\text{M}}
The number of moles of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} are, 9.898=0.1\dfrac{{9.8}}{{98}} = 0.1
The molarity of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}solution=0.1100=10−3M = \dfrac{{0.1}}{{100}} = {10^{ - 3}}{\text{M}}
40L40{{L}} of solution A will contain 4×10−24 \times {10^{ - 2}} moles of NaOH{\text{NaOH}}
10L10{{L}} of solution B will contain 10−2{10^{ - 2}} moles of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}
The equivalents of NaOH{\text{NaOH}} and H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} are 4×10−24 \times {10^{ - 2}} and 2×10−22 \times {10^{ - 2}} respectively.
After neutralization, 2×10−22 \times {10^{ - 2}} equivalents of NaOH{\text{NaOH}} remain.
[OH−]=2×10−250=4×10−4N\left[ {{\text{O}}{{\text{H}}^ - }} \right] = \dfrac{{2 \times {{10}^{ - 2}}}}{{50}} = 4 \times {10^{ - 4}}{\text{N}}
As a result we can write, [H+]=10−144×10−4=2.5×10−11N\left[ {{{\text{H}}^ + }} \right] = \dfrac{{{{10}^{ - 14}}}}{{4 \times {{10}^{ - 4}}}} = 2.5 \times {10^{ - 11}}{\text{N}}
The pH of the solution will be equal to
pH=−log⁡[H+]=−log⁡(2.5×10−11){\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right] = - \log \left( {2.5 \times {{10}^{ - 11}}} \right)
∴pH=10.6\therefore {\text{pH}} = 10.6

Note:
The n factor is also known as the valence factor and its value varies depending upon the compound being considered. For an acid, the n factor is the basicity of the acid. Similarly, the n factor of a base is its acidity. For electrolytes, it is the valency of the ions in the solution.