JEE Advanced levelSingle correct optionDifficulty 3 of 3ia033
A 0.10 M solution of the sodium salt NaHA of a diprotic acid HX2A (with pKa1=1.90) has a measured pH of 4.00 at 298 K. Assuming the amphiprotic formula applies, Ka2 of HX2A is
A1.0×10−4
B7.9×10−3
C7.9×10−7
D1.3×10−10
Answer: C
For an amphiprotic anion, pH =21(pKa1+pKa2), so pKa2=2(4.00)−1.90=6.10 and Ka2=10−6.10=7.9e−07.