Question

Question #4e55d

Answer 1

`"270 ppm Al"^(3+)`

Explanation:

The first thing to do here is convert your values to liters and grams, respectively. This will make the calculations a lot easier as we go.

So, to convert gallons to liters and pounds to kilograms, use the following conversion factors

`"1 gal " ~~ " 3.7854 L"" "` and `" " "1 kg " ~~ " 2.2046 lbs"`

To go from kilograms to Grams, use the conversion factor

`"1 kg" = 10^3"g"`

So, you will have

`660 color(red)(cancel(color(black)("gal"))) * "3.7854 L"/(1color(red)(cancel(color(black)("gal")))) = "2498.4 L"`

`9.4color(red)(cancel(color(black)("lbs"))) * (1color(red)(cancel(color(black)("kg"))))/(2.2046color(red)(cancel(color(black)("lbs")))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "4263.8 g"`

Now, the problem wants you to find the concentration of aluminium cations, `"Al"^(3+)`, expressed in parts per million, ppm. In order to find the concentration of a solute in ppm, you must essentially figure out how many grams of that solute you get in `10^6` grams of solvent.

`color(blue)(|bar(ul(color(white)(a/a)"ppm" = "grams of solute"/"grams of solvent" xx 10^6color(white)(a/a)|)))`

So, if you have `"1 g"` of solute in `10^6` grams of solvent, you have a `"1 ppm"` concentration.

Now, aluminium sulfate, `"Al"_color(red)(2)("SO"_4)_3`, contains `color(red)(2)` aluminium cations. In order to find how much aluminium you get per `"100 g"` of aluminium sulfate, calculate the compound's percent composition.

To do that, use the molar mass of aluminium and the molar mass of aluminium sulfate

`"For Al: " " " " " " "M_M = "26.98 g mol"^(-1)`

`"For Al"_2("SO"_4)_3:" " M_M = "342.15 g mol"^(-1)`

So, the percent composition of aluminium sulfate is

`(color(red)(2) xx 26.98 color(red)(cancel(color(black)("g mol"^(-1)))))/(342.15color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "15.77% Al"`

This means that `"100 g"` of aluminium sulfate will contain `"15.77 g"` of aluminium, i.e. aluminium cations. In your case, the sample of aluminium sulfate will contain

`4263.8color(red)(cancel(color(black)("g Al"_2("SO"_4)_3))) * overbrace("15.77 g Al"^(3+)/(100color(red)(cancel(color(black)("g Al"_2("SO"_4)_3)))))^(color(purple)("15.77% Al")) = "672.4 g Al"^(3+)`

To get the mass of solvent, which in your case is water, use water's density. If no information is provided, you can assume it to be equal to `"1.0 g mL"^(-1)`.

Remember that `"1 L" = 10^3"mL"`, so

`2498.4color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("water's density")) = 2.4984 * 10^6"g"`

So, you know how many grams of aluminium cations you have and how many grams of water you have, which means that you can now find the concentration in ppm

`"ppm" = (672.4 color(red)(cancel(color(black)("g"))))/(2.4984 * color(blue)(cancel(color(black)(10^6)))color(red)(cancel(color(black)("g")))) * color(blue)(cancel(color(black)(10^6))) ="269.13 ppm"`

Rounded to two sig figs, the answer will be

`"ppm Al"^(3+) = color(green)(|bar(ul(color(white)(a/a)"270 ppm"color(white)(a/a)|)))`