Question

Arrange the following resultant mixtures in increasing order of their pH values}
A. \(10 \text{ mL } 0.2 \text{ M Ca(OH)}_2 + 25 \text{ mL } 0.1 \text{ M HCl}\)
B. \(10 \text{ mL } 0.01 \text{ M H}_2\text{SO}_4 + 10 \text{ mL } 0.01 \text{ M Ca(OH)}_2\)
C. \(10 \text{ mL } 0.1 \text{ M H}_2\text{SO}_4 + 10 \text{ mL } 0.1 \text{ M KOH}\)

• A. B <C <A
• B. C <A <B
• C. C <B <A
• D. A <C <B

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question requires calculating the net acidity or basicity of three different mixtures of strong acids and strong bases to determine their relative pH order. 
Step 2: Key Formula or Approach: 
1. Calculate millimoles (mmol) of \(H^+\) and \(OH^-\) for each case. 
2. If \(H^+>OH^-\), the solution is acidic (\(pH < 7\)). 
3. If \(H^+ = OH^-\), the solution is neutral (\(pH = 7\)). 
4. If \(OH^->H^+\), the solution is basic (\(pH>7\)). 
Step 3: Detailed Explanation: 
Mixture A: 
mmol of \(Ca(OH)_2 = 10 \times 0.2 = 2 \text{ mmol}\). 
mmol of \(OH^- = 2 \times 2 = 4 \text{ mmol}\) (as \(Ca(OH)_2\) provides 2 \(OH^-\) ions). 
mmol of \(HCl = 25 \times 0.1 = 2.5 \text{ mmol}\). 
mmol of \(H^+ = 2.5 \text{ mmol}\). 
Net \(OH^- = 4 - 2.5 = 1.5 \text{ mmol}\). 
The solution is basic, so pH >7. 
Mixture B: 
mmol of \(H_2\text{SO}_4 = 10 \times 0.01 = 0.1 \text{ mmol}\). 
mmol of \(H^+ = 0.1 \times 2 = 0.2 \text{ mmol}\). 
mmol of \(Ca(OH)_2 = 10 \times 0.01 = 0.1 \text{ mmol}\). 
mmol of \(OH^- = 0.1 \times 2 = 0.2 \text{ mmol}\). 
Net \(H^+ = \text{Net } OH^- = 0\). 
The solution is neutral, so pH = 7. 
Mixture C: 
mmol of \(H_2\text{SO}_4 = 10 \times 0.1 = 1 \text{ mmol}\). 
mmol of \(H^+ = 1 \times 2 = 2 \text{ mmol}\). 
mmol of \(KOH = 10 \times 0.1 = 1 \text{ mmol}\). 
mmol of \(OH^- = 1 \text{ mmol}\). 
Net \(H^+ = 2 - 1 = 1 \text{ mmol}\). 
The solution is acidic, so pH <7. 
Ordering by increasing pH: C (Acidic) <B (Neutral) <A (Basic). 
Step 4: Final Answer: 
The correct order is C <B <A.