Answer: A, B, C
For the amphiprotic ion, pH≈(pKa1+pKa2)/2=8.34. For carbonate, Kb=10−14/10−10.33=2.14e−04, and x2/(0.10−x)=Kb gives [OHX−]=4.52e−03 M, pH =11.65. 0.10 M NaOH has pH 13, so the order holds. A pH of 8.3 is alkaline, so the last statement is false.