Asked by Kevin
A solution is prepared by adding 45.0g of Na2SO4*6H2O to pure water to make 150mL of solution. What are molar concentrations of these ions in the resulting solution?
Sodium Ion? Sulfate Ion? Phosphate Ion?
I've determined the solution is 1.2M
45/250.16g/mol Na2SO4*6H2O = .180 mol
.180 mol Na2SO4*6H2O/.150 L=1.2M
Where do I go now?
Sodium Ion? Sulfate Ion? Phosphate Ion?
I've determined the solution is 1.2M
45/250.16g/mol Na2SO4*6H2O = .180 mol
.180 mol Na2SO4*6H2O/.150 L=1.2M
Where do I go now?
Answers
Answered by
DrBob222
You've done the hard part.
(Na2SO4.6H2O) = 1.2 M.
Now there is 1 sulfate/1 Na2SO4 so (SO4^2-) must the same as (Na2SO4). Right? Right.
Now there are 2 Na/1 Na2SO4 so (Na^+) = 2 x (Na2SO4) = 2*1.2 = 2.4. Right?
(Na2SO4.6H2O) = 1.2 M.
Now there is 1 sulfate/1 Na2SO4 so (SO4^2-) must the same as (Na2SO4). Right? Right.
Now there are 2 Na/1 Na2SO4 so (Na^+) = 2 x (Na2SO4) = 2*1.2 = 2.4. Right?
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