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")#