Question

125ml of 8% w/w naoh solution(sp. Gravity 1) is added to 125ml of10% w/v hcl solution.The nature of resultant solution would be

Answer 1

Answer : The moles of acid are still present in the solution after mixing. So, the solution will become acidic is nature.

Explanation :

In 125 ml of 10% NaOH solution.

The mass of NaOH will be,

Formula used : $Mass\%=\frac{\text{Mass of NaOH}}{\text{Volume of solution}}\times 100$

$10\%=\frac{\text{Mass of NaOH}}{125ml}\times 100$

$\text{Mass of NaOH}=12.5g$

$\text{Moles of NaOH}=\frac{12.5g}{40g/mol}=0.3125mole$

In 125 ml of 10% HCl solution.

The mass of HCl will be,

Formula used : $Mass\%=\frac{\text{Mass of HCl}}{\text{Volume of solution}}\times 100$

$10\%=\frac{\text{Mass of HCl}}{125ml}\times 100$

$\text{Mass of HCl}=12.5g$

$\text{Moles of HCl}=\frac{12.5g}{36.5g/mol}=0.3424mole$

The balanced chemical reaction will be,

$NaOH+HCl\rightarrow NaCl+H_2O$

From balanced reaction we conclude that,

As, 1 mole of NaOH will neutralize 1 mole of HCl

So, 0.3125 moles will neutralize 0.3125 moles of HCl

Thus, the moles of HCl left un-neutralized = 0.3424 - 0.3125 = 0.0299 mole

That means, the moles of acid are still present in the solution after mixing. So, the solution will become acidic is nature.