Question

100ml of solution of pH = 6 is diluted to 1 litre. Resulting solution has pH is :
(A) 7.0
(B) 6.8
(C) 7.3
(D) 6.4

Answer 1

The pH is a measure of how acidic or alkaline a substance is. It runs from 0 to 14, with a pH of 7 being neutral, lower than 7 being acidic and a pH higher than 7 is alkaline. The scale is logarithmic, meaning each pH value below 7 is ten times more acidic than the next higher value.

Complete step by step answer:
The question is based on dilution. So, in order to get our answer, first, we have to understand what dilution is and how can it affect the pH of an acid or base.
To dilute an aqueous solution, we simply add water to it. This increases the proportion of the solvent or the liquid material for dilution compared to that of the solute or the component dissolved in the solvent for that given solution. For example, if you dilute saltwater, the solution will contain the same amount of salt, but the amount of water will increase hence decreasing the concentration of salt in the solution.
The effects are different in diluting the acid and base and they are as follows:
- Diluting an acid: Some examples of acidic substances include black coffee, battery acid and lemon juice. Diluting an acid decreases the concentration of H+(aq) ions, which increases the pH level of the solution and gets close to 7, making it less acidic.
- Diluting an alkali: Alkaline substances include ammonia, baking powder, caustic soda etc. Diluting an alkali decreases the concentration of OH-(aq) ions, which decreases the pH level of the solution and it also leads to 7, making it less alkaline.
Now, let's calculate the change in pH on dilution:
By using the relation , we get

& {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\\ & {{10}^{-6}}\times 100=[{{H}^{+}}]\times 1000 \\\ & [{{H}^{+}}]={{10}^{-7}} \\\ & pH=-\log ([{{10}^{-7}}]) \\\ & pH=7 \\\ \end{aligned}$$ From the above discussion, we know that the pH of an acidic solution upon dilution with water can never be equal or more than 7. In such a problem, we must consider the contribution of [${{H}^{+}}$] made by water. Let [${{H}^{+}}$] from water be x $mol{{L}^{-1}}$ in the presence of ${{10}^{-7}}$ (M) of acid. $${{\text{H}}_{\text{2}}}\text{O}\rightleftharpoons \underset{\text{1}{{\text{0}}^{\text{-7}}}\text{+x}}{\mathop{{{\text{H}}^{\text{+}}}}}\,\text{+}\underset{\text{x}}{\mathop{\text{O}{{\text{H}}^{\text{-}}}}}\,$$ At equilibrium, $$\begin{aligned} & {{\text{K}}_{\text{w}}}\text{= }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{1 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-14}}}\text{=(1}{{\text{0}}^{\text{-7}}}\text{+x)(x)} \\\ & \text{x = 0}\text{.618 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-7}}} \\\ \end{aligned}$$ Therefore, $$[{{H}^{+}}]={{10}^{-7}}+x={{10}^{-7}}+\text{0}\text{.618}\times \text{1}{{\text{0}}^{-7}}=1.\text{618}\times \text{1}{{\text{0}}^{-7}}$$ $$pH=-\log (1.618\times {{10}^{-7}})=6.79$$ **Therefore, the correct answer to the above question is an option (B), 6.8.** **Note:** Always remember that the pH level of an acidic solution cannot become greater than 7 unless the water you add to dilute it is alkaline. Similarly, the pH level of an alkaline solution cannot become lower than 7 unless the water you add to dilute it is acidic.