Animated Solution for Chemistry - Basic Concepts in Chemistry: 50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is

Imagine you are standing in a chemistry lab, staring at a classic titration setup. In the burette above, you have $50\text{ mL}$ of a $0.5\text{ M}$ oxalic acid ($H_2{C}_2{O}_4$) solution. In the conical flask below, there is $25\text{ mL}$ of a sodium hydroxide ($NaOH$) solution of unknown concentration.

Our ultimate mission is not just to find the concentration of the $NaOH$, but to determine the exact mass of $NaOH$ present in a hypothetical $50\text{ mL}$ sample of that same base solution.

Law of Equivalence

When the acid completely neutralizes the base, a beautiful chemical balance is achieved. This is governed by the Law of Equivalence, which states that at the neutralization point, the milliequivalents of the acid must exactly equal the milliequivalents of the base.

To calculate milliequivalents, we use the formula $Meq=M\times V\times n\text{-factor}$. The $n\text{-factor}$ is crucial here. Oxalic acid ($H_2{C}_2{O}_4$) is a dibasic acid, meaning it can donate two protons ($H^{+}$), so its $n\text{-factor}$ is $2$. Sodium hydroxide ($NaOH$), on the other hand, provides one hydroxide ion ($O{H}^{-}$), giving it an $n\text{-factor}$ of $1$.

Let's substitute our known values into the equivalence equation. For the acid, we multiply its molarity, volume, and $n\text{-factor}$:

Solving this is straightforward. The left side simplifies to $50$. Dividing by $25$ gives us the molarity of the sodium hydroxide solution:

We now know that the $NaOH$ solution has a concentration of exactly $2\text{ M}$.

Scaling the Volume

Here is where many students make a silly mistake. The question does not ask for the mass of $NaOH$ in the $25\text{ mL}$ we titrated. It specifically asks for the amount of $NaOH$ in a $50\text{ mL}$ sample of this solution.

We use the standard molarity formula, which relates molarity to weight ($W$), molar mass ($M_w$), and volume in milliliters ($V$):

The molar mass of $NaOH$ is $40\text{ g/mol}$. Substituting our values:

Let's calculate the weight. $1000$ divided by $50$ is $20$. And $20$ divided by $40$ is $\frac{1}{2}$.

Multiplying both sides by $2$ gives us our final weight:

The Bonus Revelation

Our calculated mass is $4\text{ g}$. But if we look at the options provided in the exam—$40\text{ g}$, $80\text{ g}$, $20\text{ g}$, and $10\text{ g}$—none of them match!

Don't panic. This was actually a famous "bonus" question in the JEE exam where all the given options were incorrect. The true answer is undeniably $4\text{ g}$. This is a powerful reminder: when you know your concepts are rock solid and your calculations are flawless, trust your math over the printed options!