JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia110
A 0.10M solution of ammonium formate, HCOONHX4, is prepared. Take Ka(HCOOH)=1.8×10−4 (pKa=3.75), Kb(NHX3)=1.8×10−5 (pKb=4.75) and Kw=1.0×10−14. Use pH=7+21(pKa−pKb) and h=Kw/(KaKb). Select the correct statement(s).
AThe pH of the solution is 6.50
BThe pH changes appreciably if the solution is diluted to 0.010M
CThe degree of hydrolysis of the salt is about 1.8×10−3
D[NHX3] and [HCOOH] are nearly equal, each about 1.8×10−4M
Answer: A, C, D
pH=7+21(3.75−4.75)=6.50, independent of concentration, so the second statement is false. h=10−14/(1.8×10−4×1.8×10−5)=1.8×10−3. Hydrolysis produces NHX3 and HCOOH in nearly equal amounts: 0.10×h=1.8×10−4M each.