question below (under details)?

0 ⤊

question below (under details)?

Starting with 14.0 ml of 3.50 × 10^−2 M AgNO3, how many grams of KBr can be added to the solution before AgBr begins to precipitate ?

2007-03-24 20:52:20 · 2 answers · asked by meister 2 in Science & Mathematics ➔ Chemistry

Ksp of AgBr= 3.3 X 10^ -13

2007-03-24 21:45:32 · update #1

2 answers

looks like there's no Br- ions could escape from Ag+ ions.. So the answer according to me is zero......

peace
vixklen

2007-03-24 20:57:44 · answer #1 · answered by vixklen 3 · 0⤊ 0⤋

Molar mass of KBr = 39+80 = 119 g

Thus grams of KBr = 14x3.50x10^-2x10^-3x119 g = 0.05831 g

That's to ppt out all the Ag, however, Ksp of AgBr is reqd.

2007-03-24 21:27:32 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋