A solution of Na2HPO4 in excess is treated with 0.030 mol of AgNO3, forming yellow Ag3PO4 completely according to 3Ag+ + HPO4^2- -> Ag3PO4 + H+. The amount of H+ (as free acid, HNO3) liberated and the effect on the medium are

0.010 mol; the solution becomes acidic

0.020 mol; the solution stays neutral

0.030 mol; the solution becomes acidic

0.010 mol; the solution becomes more alkaline

0.010 mol; the solution becomes acidic

JEE Advanced levelSingle correct optionDifficulty 3 of 3qa203

A solution of $Na X_{2} HPO X_{4}$ in excess is treated with 0.030 mol of $AgNO X_{3}$ , forming yellow $Ag X_{3} PO X_{4}$ completely according to $3 Ag X^{+} + HPO X_{4} X^{2 -} Ag X_{3} PO X_{4} + H X^{+}$ . The amount of $H X^{+}$ (as free acid, $HNO X_{3}$ ) liberated and the effect on the medium are

HPO X_{4} X^{2 -}

must lose one proton to supply

PO X_{4} X^{3 -}

, so each

Ag X_{3} PO X_{4}

releases one

H X^{+}

. The

Ag X^{+}

is 0.030 mol, giving

0.030/3 = 0.010

mol of

Ag X_{3} PO X_{4}

and 0.010 mol of

H X^{+}

. The medium turns acidic, which is why a buffer or ammonia is used to complete the precipitation.