Question #4e55d

1 Answer
Mar 10, 2016

Answer:

#"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)|)))#