Animated Solution for Chemistry - Ionic Equilibrium: Two solutions, $A$ and $B$, each of $100\text{ L}$ was made by dissolving $4\text{ g}$ of $\text{NaOH}$ and $9.8\text{ g}$ of ${\text{H}}_2{\text{SO}}_4$ in water, respectively. The pH of the resultant solution obtained from mixing $40\text{ L}$ of solution $A$ and $10\text{ L}$ of solution $B$ is ……… .

The problem of mixing two different solutions—one a strong base and the other a strong acid—is a classic test of your understanding of stoichiometry and ionic equilibrium. It’s not just about plugging numbers into a formula; it’s about visualizing the physical reality of molecules interacting in a massive tank. Let's break down this epic clash of acids and bases step by step.

Imagine you are standing in front of two massive tanks. Tank A holds $100\text{ L}$ of a sodium hydroxide ($\text{NaOH}$) solution, and Tank B holds $100\text{ L}$ of a sulfuric acid (${\text{H}}_2{\text{SO}}_4$) solution.

Before we can even think about mixing them, we need to know exactly how concentrated each solution is. We are given the mass of the solutes: $4\text{ g}$ of $\text{NaOH}$ and $9.8\text{ g}$ of ${\text{H}}_2{\text{SO}}_4$.

To find the molarity, we first convert these masses into moles. For $\text{NaOH}$, the molar mass is $40\text{ g/mol}$. Dividing this by the total volume of $100\text{ L}$ gives us the molarity of Solution A:

Now, let's look at Solution B. The molar mass of ${\text{H}}_2{\text{SO}}_4$ is $98\text{ g/mol}$. Similarly, dividing by $100\text{ L}$ gives us the molarity of Solution B:

Now the real game begins. We aren't mixing the entire tanks. We are carefully extracting $40\text{ L}$ from Tank A and $10\text{ L}$ from Tank B, and pouring them into a brand new mixing container.

First, let's calculate the exact number of moles of hydroxide ions (${\text{OH}}^{-}$) coming from Solution A. Since $\text{NaOH}$ is a strong base, it dissociates completely.

Next, we calculate the moles of hydrogen ions (${\text{H}}^{+}$) coming from Solution B. Here is where many students make a silly mistake! Sulfuric acid is a strong dibasic acid. This means every single molecule of ${\text{H}}_2{\text{SO}}_4$ releases two ${\text{H}}^{+}$ ions into the solution.

When these two solutions meet in the mixing tank, a fierce neutralization reaction occurs. Every ${\text{H}}^{+}$ ion seeks out an ${\text{OH}}^{-}$ ion to form water.

Comparing the moles, we have $40\times 10^{-3}\text{ mol}$ of ${\text{OH}}^{-}$ and only $20\times 10^{-3}\text{ mol}$ of ${\text{H}}^{+}$. The hydroxide ions are clearly in excess! The ${\text{H}}^{+}$ ions will be completely consumed, leaving behind a surplus of ${\text{OH}}^{-}$.

To find the final concentration of these excess hydroxide ions, we must divide by the new total volume of the mixture. We mixed $40\text{ L}$ and $10\text{ L}$, so the total volume is $50\text{ L}$.

With the hydroxide concentration in hand, we can easily find the pOH. Given that $\log 4\approx 0.602$, we get:

Finally, at standard room temperature, the relationship between pH and pOH is $\text{pH}+\text{pOH}=14$.

Rounding to two decimal places, we arrive at our final answer: 10.60. The solution is highly basic, exactly as we expected from the excess of hydroxide ions!