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

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")#