Question

The \(K_a\) values of A, B and C are \(1.8\times 10^{-4}\), \(5\times 10^{-10}\) and \(3\times 10^{-8}\) respectively. The correct order of their acidic strength is

• A. \(B\gt A\gt C\)
• B. \(B\gt C\gt A\)
• C. \(A\gt B\gt C\)
• D. \(A\gt C\gt B\)

Hint:

Acidic strength is directly proportional to the acid dissociation constant: \[ \text{Higher }K_a \Rightarrow \text{stronger acid} \] So, compare \(K_a\) values directly to arrange acidic strength.

Answer 1

The Correct Option is D

Step 1: Recall the relationship between $K_a$ and acid strength.
A higher $K_a$ value means a stronger acid. The acid dissociation constant $K_a$ directly measures the extent of ionization: larger $K_a$ = more ionization = stronger acid.
Step 2: List the given $K_a$ values.
Acid A: $K_a = 1.8 \times 10^{-4}$. Acid B: $K_a = 5 \times 10^{-10}$. Acid C: $K_a = 3 \times 10^{-8}$.
Step 3: Convert to comparable form.
A: $1.8 \times 10^{-4} = 18,000 \times 10^{-8}$. C: $3 \times 10^{-8}$. B: $0.05 \times 10^{-8}$. Clearly A >> C >> B when expressed with the same power of 10.
Step 4: Rank the acids from strongest to weakest.
$K_a(A) = 1.8 \times 10^{-4} > K_a(C) = 3 \times 10^{-8} > K_a(B) = 5 \times 10^{-10}$. Therefore: A > C > B in acidic strength.
Step 5: Verify the order makes sense.
pKa values: pKa(A) = $-\log(1.8 \times 10^{-4}) \approx 3.74$; pKa(C) = $-\log(3 \times 10^{-8}) \approx 7.52$; pKa(B) = $-\log(5 \times 10^{-10}) \approx 9.30$. Lower pKa = stronger acid. Order: A (pKa 3.74) > C (pKa 7.52) > B (pKa 9.30). Confirmed.
Step 6: State the final answer.
The correct order of decreasing acidic strength is A > C > B.
\[ \boxed{A > C > B} \]