Question

What is the pH of resulting solution when \(20\text{ mL }\frac{M}{10}\text{ NaOH}\) and \(10\text{ mL }\frac{M}{10}\text{ H}_2\text{SO}_4\) are mixed together ?

• A. \(0\)
• B. \(2\)
• C. \(7\)
• D. \(10\)

Hint:

Compare the moles of OH- with the moles of H+ before finding pH.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
NaOH is a monoacidic base and \(\text{H}_2\text{SO}_4\) is a dibasic acid. We must compare the moles of \(\text{OH}^-\) with those of \(\text{H}^+\).

Step 2: Moles of each:
Moles of NaOH = \(20\times 0.1 = 2\) millimol, so \(\text{OH}^-\) = 2 millimol.
Moles of \(\text{H}_2\text{SO}_4\) = \(10\times 0.1 = 1\) millimol, giving \(\text{H}^+\) = 2 millimol.

Step 3: Neutralisation:
\[ 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \]
The \(\text{OH}^-\) and \(\text{H}^+\) are exactly equal, so both are used up completely. The resulting solution is sodium sulphate, a salt of a strong acid and strong base, which does not hydrolyse. So \([\text{H}^+] = 10^{-7}\) and pH = 7.
pH 0, 2 or 10 would need leftover acid or base, which does not happen here.

Final Answer:
The solution is neutral, so pH = 7, option (C). \[ \boxed{\text{pH} = 7} \]