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
The Ksp for AgBrO3 is 4.00 x 10^-2
I got 9.43 x 10^-2 g
1 Answer
I got
Well, if we assume that the volume of the solution does not change, then we begin with
"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) = 4.00 xx 10^(-2) = s(0.500 + s)
This
s^2 + 0.500s - 0.0400 = 0
This quadratic equation gives
So, in
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
0.05 xx "16.6 g" = color(blue)ul("0.827 g")