Problem 250 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate the \(\mathrm{pH}\) of \(10^{-8} \mathrm{M} \mathrm{HCl} \cdot(\log 11=1.0414)\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 6.96.

Step 1: Understand the Problem

We are asked to calculate the \(\mathrm{pH}\) of a \(10^{-8} \mathrm{M}\) solution of \(\mathrm{HCl}\), an acid. When an acid is dissolved in water, it dissociates into hydrogen ions. \(\mathrm{HCl}\) is a strong acid, so it will fully dissociate into \(\mathrm{H}^+\) ions and \(\mathrm{Cl}^-\) ions. However, because the concentration is very low, it gets complicated by the autoionization of water.

Step 2: Consider the Contribution of Water

The solution's hydrogen ion concentration isn’t just from the \(\mathrm{HCl}\); water autoionization contributes \([\mathrm{H}^+] = 10^{-7} \mathrm{M}\). Because \([\mathrm{HCl}] = 10^{-8} \mathrm{M}\) and is lower than this, we should add these concentrations.

Step 3: Calculate Total Hydrogen Ion Concentration

Total \([\mathrm{H}^+] = 10^{-8} \mathrm{M} + 10^{-7} \mathrm{M}\). This gives us \([\mathrm{H}^+]=1.1 \times 10^{-7} \mathrm{M}\).

Step 4: Calculate the pH

The formula for \(\mathrm{pH}\) is \(\mathrm{pH} = -\log_{10}([\mathrm{H}^+])\). Substitute \([\mathrm{H}^+]=1.1 \times 10^{-7}\) to find \(\mathrm{pH} = -\log(1.1 \times 10^{-7})\). Use the reference \(\log 11 = 1.0414\).

Step 5: Resolve the Logarithmic Expression

Rewrite \(\log(1.1 \times 10^{-7})\) as \(\log(1.1) + \log(10^{-7})\). We know \(\log(1.1) = 0.0414\) (from \(\log 11 = 1.0414\)) and \(\log(10^{-7}) = -7\). Thus, \(\log(1.1 \times 10^{-7}) = 0.0414 - 7\), which equals \(-6.9586\).

Step 6: Find the Final pH

Calculate \(\mathrm{pH} = -(-6.9586) = 6.9586\). Thus the \(\mathrm{pH}\) of the solution is approximately \(6.96\).

Key Concepts

Autoionization of Water

Water can actually generate ions on its own, even when no acid or base is added. Isn't that interesting? This process is called autoionization. In pure water, a small fraction of water molecules will dissociate into hydrogen ions (H^+) and hydroxide ions (OH^-). The equation representing this reaction is:

\[ 2 \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]
For simplicity, we often refer to hydrogen ions as just H^+. Even though water ionizes only slightly, it is enough to have a measurable effect on acidic or basic solutions. In pure water at 25^{\circ}C, \[[\text{H}^+] = [\text{OH}^-] = 10^{-7} \text{M}\], giving a neutral pH of 7. If a strong acid or base is added to water, these values will change, but the autoionization still makes a contribution.

Strong Acids

Strong acids dissociate completely in water, which means they break apart into their ions. Take hydrochloric acid (HCl) as an example. When it is mixed with water, it separates into H^+ ions and Cl^- ions almost entirely:

\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Because of this complete dissociation, the concentration of hydrogen ions in the solution will directly come from the acid concentration. Now, here's the catch. In very dilute solutions, such as 10^{-8} \text{M} HCl, the hydrogen ions aren't just coming from the acid – the autoionization of water contributes significantly too. This is why, for our problem, calculating total H^+ concentration involves adding hydrogen ions from both HCl and water.

Hydrogen Ion Concentration

The hydrogen ion concentration is a crucial player in determining the pH of a solution. pH is calculated using:
\[ \text{pH} = -\log_{10}([\text{H}^+]) \]
So, what affects [H^+]? For strong acids like HCl, it is straightforward: its concentration gives [H^+] as it dissociates completely. But if that concentration is lower than 10^{-7} \text{M} (as with our given 10^{-8} \text{M} HCl solution), the water's autoionization contribution of 10^{-7} \text{M} takes precedence. Consequently, the total H^+ concentration is a sum of the contributions from the acid itself and the water, showing that water can have a noticeable role in dilute solutions. After finding the total H^+ concentration, we can finally determine the pH by substitution into the formula.