Question

$125mL$ of $63$% $\left(w/v\right)H_2{C}_2{O}_4{2H}_{2}O$ solution is made to react with $125mL$ of a $40$% $\left(w/v\right)NaOH$ solution.the resulting solution is

• A. neutral
• B. acidic
• C. strongly acidic
• D. alkaline

Answer 1

The correct option is A neutral

We have,

$125mL$ of $63\%$ $\left(w/v\right)$ $H_2{C}_2{O}_4.2H_{2}O$$=\frac{63\times 125}{100}g$

$i.e$

$=78.75g=$$\frac{78.75}{63}\text{gram equivalent oxalic acid}=1.25\text{g-equiv oxalic acid.}$

$Also,$

$125mL$ of $40\%$ $\left(w/v\right)NaOH$ solution

$i.e$

$=\frac{40\times 125}{100}$g of $NaOH$$=50gNaOH=$$\frac{50}{40}$g-equiv $NaOH$$=1.25gram-equivalent$ $NaOH$

According to the law of equivalence, the equivalent amount of base reacts with an equivalent amount of acid to form an equivalent amount of salt. $\text{Here the equivalent amount of base and acid are equal.}$So the resultant solution will be $neutral.$

Option A is correct.