Question

125 ml of 10% NaOH (w/V) is added to 125ml of ${\text{10}}$% ${\text{HCl}}$ (w/V). The resultant solution becomes
A. alkaline
B. strongly alkaline
C. Acidic
D. neutral

Answer 1

To determine the nature of the solution we will determine the number of moles of each. After writing the balanced equation, by comparing the number of moles of reactant and product, the amount of left reactant can be determined. The nature of the left reactant will decide the nature of the solution.

Formula used:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete answer:
${\text{125}}\,{\text{ml}}\,$ of ${\text{10}}$% ${\text{NaOH}}$(w/V) is added to${\text{125}}\,{\text{ml}}\,$ of ${\text{10}}$% ${\text{HCl}}$(w/V).
${\text{10}}$% ${\text{NaOH}}$(w/V) means $10\,{\text{g}}$ ${\text{NaOH}}$ is added in $100\,{\text{ml}}$ solution. So, the gram amount of ${\text{NaOH}}$ present in ${\text{125}}\,{\text{ml}}\,$ solution is,
$\,100\,{\text{ml}}\, = \,10\,{\text{gNaOH}}\,$
$\Rightarrow 125\,{\text{ml}}\, = \,12.5\,{\text{g}}\,\,{\text{NaOH}}\,$
So, $12.5\,{\text{g}}$ ${\text{NaOH}}$ is present in ${\text{125}}\,{\text{ml}}\,$ solution.
${\text{10}}$% ${\text{HCl}}$ (w/V) means $10\,{\text{g}}$ ${\text{HCl}}$ is added in $100\,{\text{ml}}$ solution. So, the gram amount of ${\text{HCl}}$ present in ${\text{125}}\,{\text{ml}}\,$ solution is,
$\,100\,{\text{ml}}\, = \,10\,{\text{g}}\,{\text{HCl}}\,$
$\Rightarrow \,125\,{\text{ml}}\, = \,12.5\,{\text{g}}\,\,{\text{HCl}}\,$
So, $12.5\,{\text{g}}$ ${\text{HCl}}$ is present in ${\text{125}}\,{\text{ml}}\,$ solution.
We will use the mole formula to determine the mole of ${\text{NaOH}}$ and ${\text{HCl}}$ present in $12.5\,{\text{g}}$ follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of ${\text{NaOH}}$ is $40\,{\text{g/mol}}$.
Substitute $40\,{\text{g/mol}}$ for molar mass and $12.5\,{\text{g}}$ for moles for ${\text{NaOH}}$.
$\Rightarrow {\text{mole}}\,\,{\text{ = }}\,\dfrac{{{\text{12}}{\text{.5}}\,{\text{g}}}}{{40\,{\text{g/mol}}}}$
$\Rightarrow {\text{mole}}\,{\text{ = }}\,{\text{0}}{\text{.3125}}$
So, the mole of ${\text{NaOH}}$ present in $12.5\,{\text{g}}$is ${\text{0}}{\text{.3125}}$.
Molar mass of ${\text{HCl}}$ is $36.5\,{\text{g/mol}}$.
Substitute $36.5\,{\text{g/mol}}$ for molar mass and $12.5\,{\text{g}}$ for moles of ${\text{HCl}}$.
$\Rightarrow {\text{mole}}\,\,{\text{ = }}\,\dfrac{{{\text{12}}{\text{.5}}\,{\text{g}}}}{{36.5\,{\text{g/mol}}}}$
$\Rightarrow{\text{mole}}\,{\text{ = }}\,{\text{0}}{\text{.3424}}$
So, the mole of ${\text{HCl}}$ present in $12.5\,{\text{g}}$ is ${\text{0}}{\text{.3424}}$.
The sodium hydroxide reacts with hydrochloric acid and forms salt and water.
The reaction is as follows:
${\text{NaOH}}\,{\text{ + }}\,{\text{HCl}} \to {\text{NaCl + }}\,{{\text{H}}_{\text{2}}}{\text{O}}\,$
According to the balanced reaction, one mole of sodium hydroxide reacts with one mole of hydrochloric acid.
Compare the mole ratio to determine the reactant left as follows:
${\text{1}}\,{\text{mol}}\,\,{\text{NaOH = }}\,\,{\text{1}}\,{\text{mol}}\,\,{\text{HCl}}\,$
$\Rightarrow {\text{0}}{\text{.3125}}\,{\text{mol}}\,\,{\text{NaOH = }}\,\,{\text{0}}{\text{.3125}}\,{\text{mol}}\,\,{\text{HCl}}\,$
So, ${\text{0}}{\text{.3125}}$ mole o f${\text{NaOH}}$ will react with ${\text{0}}{\text{.3125}}$ mole ${\text{HCl}}$.
So, the left amount of hydrochloric acid is,
${\text{0}}{\text{.3424}}\, - 0.3125\, = \,0.0299$
So, the left amount of hydrochloric acid is $0.0299$.
So, after the completion of reaction $0.0299$ mole of acid will remain in the solution so, the solution will be acidic.

**Therefore, option (C) acidic is correct.

Note:**
The reaction of strong acid and strong base gives salt and water. The resultant solution becomes natural if acid and base both are present in the same number of moles. If acid and base present in a different number of moles then the nature of the solution will be decided based on the left species. To determine the stoichiometry relations a balanced equation is necessary.