What mass of AgBrO3 can be dissolved in 50.0 mL of 0.500M NaBrO3 at 80°C?

The Ksp for AgBrO3 is 4.00 x 10^-2

I got 9.43 x 10^-2 g

1 Answer
Jan 17, 2018

I got "0.827 g AgBrO"_3(s). The K_(sp) is rather large... so the mass that can be dissolved should be kinda big.


Well, if we assume that the volume of the solution does not change, then we begin with ["BrO"_3^(-)]_i = "0.500 M" in solution already. You should recall this as the common ion effect... which decreases solubility in the presence of the same ion.

"AgBrO"_3(s) rightleftharpoons "Ag"^(+)(aq) + "BrO"_3^(-)(aq)

"I"" "-" "" "" "" "" ""0 M"" "" "" ""0.500 M"
"C"" "-" "" "" "" "+s " M"" "" "+s " M"
"E"" "-" "" "" "" "" "s " M"" "" "(0.500 + s)"M"

Therefore, the K_(sp) expression is:

K_(sp) = 4.00 xx 10^(-2) = s(0.500 + s)

This K_(sp) is too big to make the small s approximation. We rearrange this to get:

s^2 + 0.500s - 0.0400 = 0

This quadratic equation gives s = "0.0702 M" as the only physically reasonable answer. If we had made the small s approximation, we would have gotten "0.0800 M", which would have given 14.0% error.

So, in "1 L" of solution, we expect to be able to dissolve "0.0702 mols" of "Ag"^(+), which is 1:1 with "AgBrO"_3.

This gives that this many grams of solid can be dissolved in one liter...

0.0702 cancel("mols AgBrO"_3(s)) xx "235.77 g AgBrO"_3/cancel"1 mol"

= "16.6 g AgBrO"_3(s)

Since we have "50.0 mL", we have 5% of one liter, and thus, we can dissolve

0.05 xx "16.6 g" = color(blue)ul("0.827 g")