Animated Solution for Chemistry - Basic Concepts in Chemistry: 25 mL of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2 M aqueous NaOH solution?

The beauty of volumetric analysis lies in its elegant simplicity. It is like a chemical dance where molecules pair up perfectly, and if we know the steps of one partner, we can deduce everything about the other. In this problem, we are dealing with a classic double titration scenario. We have a single bottle of hydrochloric acid ($\text{HCl}$), but its concentration is a mystery. To solve this, we will perform two distinct titrations.

Unmasking the Unknown Acid

Imagine you are in the laboratory. You take $25\text{ mL}$ of the unknown $\text{HCl}$ solution. To find its concentration, you titrate it against a standard solution of sodium carbonate (${\text{Na}}_2{\text{CO}}_3$). We are given that it takes $30\text{ mL}$ of $0.1\text{ M}$ ${\text{Na}}_2{\text{CO}}_3$ to reach the equivalence point.

The golden rule of any titration is the Principle of Equivalence. At the exact moment of neutralization, the number of equivalents of the acid must perfectly equal the number of equivalents of the base.

Mathematically, this is expressed as:

Remember that normality ($N$) is simply molarity ($M$) multiplied by the $n$-factor. For $\text{HCl}$, it releases one $H^{+}$ ion, so its $n$-factor is $1$. For ${\text{Na}}_2{\text{CO}}_3$, it can accept two $H^{+}$ ions to form carbonic acid, so its $n$-factor is $2$.

Let's set up our equivalence equation for the first titration:

Now, we perform the atomic computation. The right side simplifies beautifully: $0.1\times 2=0.2$, and $0.2\times 30=6$.

We have successfully unmasked the concentration of our acid!

The Final Showdown

Now that we know our $\text{HCl}$ has a molarity of $\frac{6}{25}\text{ M}$, we move to the second part of the experiment. We need to find out how much of this exact same acid is required to neutralize $30\text{ mL}$ of a $0.2\text{ M}$ sodium hydroxide ($\text{NaOH}$) solution.

Once again, we call upon the Principle of Equivalence. This time, our base is $\text{NaOH}$, which provides one $O{H}^{-}$ ion, giving it an $n$-factor of $1$.

Setting up the equation for the second titration:

Substitute the values we know:

The right side again simplifies to $6$.

Notice how elegantly the math unfolds. The $6$ on both sides cancels out perfectly, leaving us with:

Equivalence Transitivity

If you look closely at the math, you might notice a brilliant shortcut. In both titrations, the number of milli-equivalents of the base was exactly $6$.

Since the same acid is neutralizing both bases, and the equivalents of the bases are identical, the volume of the acid required must be identical to the volume used in the first titration! Recognizing these patterns can save you precious minutes in competitive exams like JEE.