Question

There are two different solutions, A and B. The pH of solution A is 4. Find the pH of solution B having \([\text{H}^+]\) three times that of solution A.

• A. \(3.5229\)
• B. \(4.229\)
• C. \(3.229\)
• D. \(3.5223\)

Hint:

Use pH = -log[H+] and note that tripling [H+] subtracts log 3 from the pH.

Answer 1

The Correct Option is A

Step 1: Approach
Compare the two solutions through the difference in pH rather than computing each one.

Step 2: Difference
\[ \text{pH}_A-\text{pH}_B=\log\frac{[\text{H}^+]_B}{[\text{H}^+]_A}=\log3=0.4771 \]

Step 3: Result
\[ \text{pH}_B=4-0.4771=3.5229 \]

Step 4: Sense check
A tenfold rise in $[\text{H}^+]$ would lower pH by exactly 1. Tripling is a smaller rise, so the drop must be less than 1, and it is 0.4771. The answer lies between 3 and 4, matching option (A).

Final Answer:
The pH of solution B is 3.5229, option (A). \[ \boxed{3.5229} \]