Question

125 mL of 10% NaOH (w/V) is added to 125 ml of 10% HCl (w/V)the resulting solution becomes????????1.Alkaline 2.Strongly alkaline 3. Acidic4. Neutral Plz explain the answer

Answer 1

Answer:

Moles of acid are still present in the solution after mixing trhe solution will become acidic is nature.

Explanation:

In 125 mL of 10% NaOH solution.

Mass of  NaOH in  10% NaOH (w/V) solution.

$10\%=\frac{m}{125 }\times 100$

m = 12.5 g

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

In 125 mL of 10% HCl solution.

Mass of  HCl in  10% HCl(w/V) solution.

$10\%=\frac{m'}{125 }\times 100$

m' = 12.5 g

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

$NaOH+HCl\rightarrow H_2O+NaCl$

So, 0.3125 moles will neutralize 0.3125 moles of HCl.

So, Moles of HCL left un-neutralized:

=0.3424 mol - 0.3125 moles=0.0299 mol

Since moles of acid are still present in the solution after mixing trhe solution will become acidic is nature.