Question

Which of the following salt solutions will be the most basic when equal concentrations of these salt solutions are prepared? O(A) NaBr O(B) KOBY O(C) KBr (D) KOF O(E) NH4CI Submit

Answer 1

can the strength of The anions of the given sauts are conjugate bases of acids. So compar ing the strength of the acids, we compare the bases we can eliminate, NHU Cl, NaBr, kBr. from comparison becaus NH4Cl is an acidic salt NHut solution forms acidic Solcetion. Nabr, K Br are nuetral Salts. Because is formed by reaction of strong and and strong bose: so less reactive their conjugale is very OBV is the conjugate base of HOBO (hypobromous acid KOBY - ktor et is KOBY Br + H2O - HO Brt OH ID (КО Е is conjugate base HOF (hypo flourous acid) AOF -> kt t of OF + H20 2 HOF & OH

o H F H Br IS los both acids balogen (f and Br attached to Oxygen, The higher the electronegativity of halogen, the higher the electron pall from and higher the weakening on 14 bond A strong acid is the one which loses 147 easily. Flourine has bögher electronegativity the bromine. So HOF is stronger acid than HoBr. The stronger the acid, the weaker is ets conjugate base. weaker base compared to So, for equal concentration of satts KoBr will be most basic. of is Obr.