[peachgoat950]

One more question from my assignment that I am TOTALLY lost on - again I don't want the answer; just advice on how to approach it:

How many grams of barium nitrate are required to precipitate the entire sulfate ion present in 34.1mL of 0.149M sulfuric acid solution?

Is this the same as neutralization??? Any advice is GREATLY appreciated - thanks!!!!!!!

[Astrokel]

Start with balanced equation and work with moles.

[peachgoat950]

Ok; this is what I have:

mwBa(NO3) = 199.32g/mol
mwH2SO4 = 97.07g/mol

Ba(NO3) + H2SO4 -> BaSO4 + H2(NO3)
V of H2SO4 = 34.1mL or 0.034L
M = 0.149

n=mxv = 0.149 x 0.0341 = 5.08 x 10 -3 mol
5.08 x 10 -3 mol H2SO4 x 1 mol Ba(NO3)/1 mol H2 SO4 = 5.08 x 10 -3 mol
5.08 x10 -3 mol Ba(NO3) x 199.32 g/mol/1 mol Ba(NO3) = 1.01g

1.01 g is required...I think???

[Astrokel]

Your steps took are correct except that your equation is wrong. What is the charge on nitrate?

[peachgoat950]

Oops...the charge on nitrate is -1, meaning that barium nitrate should be Ba(NO3)2, not Ba(NO3)...so the equation should be :

Ba(NO3)2 + H2SO4 -> BaSO4 + H2(NO3)2

[Astrokel]

H2(NO3)2

[peachgoat950]

 ...I'm guessing that was completely WRONG. My brain was still asleep; let me try again....

The formula for nitric acid is HNO3, so the equation should look like this:

Ba(NO3)2 + H2SO4 --> BaSO4 + 2HNO3

[AWK]

mwBa(NO3) = 199.32g/mol
mwH2SO4 = 97.07g/mol

[peachgoat950]

hmph...

mw Ba(NO3)2 - 261.34 g/mol
mw H2SO4 - 98.07 g/mol

so 5.08 x 10 -3 x 261.34 = 1.38g

[Astrokel]

so 5.08 x 10 -3 x 261.34 = 1.38g

Your steps are correct but i did not check your M_r and values.