Starting with 13.2 mL of 3.90 × 10^ −2 M AgNO3, how many grams of KBr can be added to the solution?

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Starting with 13.2 mL of 3.90 × 10^ −2 M AgNO3, how many grams of KBr can be added to the solution?

before AgBr begins to precipitate ?

2007-03-24 04:59:12 · 4 answers · asked by Paul C 1 in Science & Mathematics ➔ Chemistry

4 answers

Ksp (AgBr)= 5 10 ^-13

5 10^-13 = 3.90 10^-2 (Br-)

(Br-)= 1.28 10 ^-11 mole/l

1.28 10^-11 mole : 1000 mL = x :13.2

x= 1.69 10^-13 mole KBr

MM(KBr)= 119 g/mol

1.69 10^-13 ( 119)= 2.0 10^-11

2007-03-24 05:28:13 · answer #1 · answered by Anonymous · 1⤊ 0⤋

thats a very small ammount are you sure Ksp is right

2007-03-24 14:03:22 · answer #2 · answered by tony200423man 2 · 0⤊ 0⤋

use the ksp 3.3e-13

2007-03-26 12:31:02 · answer #3 · answered by akin k 1 · 0⤊ 0⤋

it's correct

2007-03-26 14:21:07 · answer #4 · answered by Raki 3 · 0⤊ 0⤋