Question

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

Answer 1

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.

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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")`