Animated Solution for Chemistry - Ionic Equilibrium: At $25^{\circ }\text{C}$, the concentration of ${\text{H}}^{+}$ ions in $1.00\times 10^{-3}\text{ M}$ aqueous solution of a weak monobasic acid having acid dissociation constant $(K_a)=4.00\times 10^{-11}$ is $X\times 10^{-7}\text{ M}$. The value of $X$ is _______. Use: Ionic product of water $(K_w)=1.00\times 10^{-14}$ at $25^{\circ }\text{C}$

Welcome to a classic trap in Ionic Equilibrium! When dealing with weak acids, our first instinct is often to blindly apply the standard approximation formula. But in the world of competitive exams like JEE, taking things at face value can lead to silly mistakes. Let's dive into a problem where water refuses to be ignored.

We are given a weak monobasic acid, let's call it $\text{HX}$, with a concentration $C=1.00\times 10^{-3}\text{ M}$ and an incredibly small acid dissociation constant $K_a=4.00\times 10^{-11}$.

Normally, we calculate the hydrogen ion concentration using the simplified formula:

Let's plug in our values and see what happens:

Wait a minute! The $[{\text{H}}^{+}]$ produced by the acid is $2\times 10^{-7}\text{ M}$. This is dangerously close to the $[{\text{H}}^{+}]$ naturally present in pure water at $25^{\circ }\text{C}$, which is $10^{-7}\text{ M}$. Whenever the acid's contribution is comparable to $10^{-7}\text{ M}$ (usually between $10^{-6}$ and $10^{-8}$), we cannot ignore the self-ionization of water.

Because we can't ignore water, we now have two sources of ${\text{H}}^{+}$ ions in our beaker. This is a classic case of simultaneous equilibrium governed by the common ion effect.

Let's define our variables: Let $x$ be the amount of ${\text{H}}^{+}$ contributed by the acid $\text{HX}$. Let $y$ be the amount of ${\text{H}}^{+}$ contributed by the self-ionization of ${\text{H}}_2\text{O}$.

Since both reactions occur in the same solution, the total concentration of hydrogen ions will be the sum of both contributions:

Now, let's write the equilibrium expressions for both processes. For the weak acid:

Since $K_a$ is extremely small, the dissociation $x$ is negligible compared to the initial concentration $10^{-3}\text{ M}$. We can safely approximate the denominator as $10^{-3}$. This gives us our first working equation:

For the self-ionization of water, the ionic product $K_w$ is:

We now have a system of two equations. You might be tempted to substitute $y$ from equation (2) into equation (1), but that would lead to a messy cubic equation. Instead, let's look for mathematical elegance.

Notice that both equations share the common term $(x+y)$. What happens if we simply add the two equations together?

Factor out the common $(x+y)$ term on the left side:

Isn't that beautiful? By adding the equations, we directly formed a perfect square of the exact quantity we are looking for: the total hydrogen ion concentration!

Now, all that's left is to take the square root of both sides to find the total $[{\text{H}}^{+}]$:

We know that $\sqrt{5}\approx 2.236$. Therefore:

The question asks for the answer in the format $X\times 10^{-7}\text{ M}$. Comparing our result, we find:

This problem is a fantastic reminder that chemistry is not just about memorizing formulas; it's about understanding the physical reality of the solution and applying math elegantly to describe it.