Question

It is found that 0.1 (M) solution of three sodium salts NaA, NaB and NaC have pHs 8, 9 and 11 respectively. Therefore the order of acidic strengths of HA, HB and HC are

• A. HA > HB > HC
• B. HB > HA > HC
• C. HB > HC > HA
• D. HA > HC > HB

Answer 1

Answer: (A)

HA > HB > HC

Answer 2

pH = 7 + $\frac{1}{2}$ pK_a + log c = 6 + $\frac{1}{2}$pK_a since c = 0.1(M)

for NaA, 8 = 6 + $\frac{1}{2}$ pK_a or pK_a = 4

for NaB, 9 = 6 + $\frac{1}{2}$pK_a or pK_a = 6

for NaC, 11 = 6 + $\frac{1}{2}$ pK_a or pK_a = 10