Yes, you would have concn BUT M = moles/L so if you know moles and M you can calculate L and convert to mL. I think you have the right idea; you just quit too soon.
moles NH3 = 1.74/17.03 = ?
moles (NH4)2SO4 = 4.19/132.14 = ? and that x 2 = about 0.0628
...........NH3 + H^+ ==> NH4^+
initial...0.102...0.....0.0628
add...............x
change.....-x....-x.......+x
equil.....0.102-x..0.....0.0628+x
pH = pKa + log[(base)/(acid)]
9.00 = 9.25 + log [(0.102-x)/(0.0628+x)]
Solve for x. If I didn't goof that should be close to 0.0427 mols H^+.
M = moles/L
M = 14, moles = 0.0427, solve for L and convert to mL and I get approximately 3 mL but you need to be a little more accurate than that. I always like to check it to see if I get the right pH.
NH3 = 0.102-0.0427 = 0.0593
NH4^+ = 0.0628+0.0427 = 0.1055
pH = 9.25 + log (0.0593/0.1055) = 9.00
Voila.
You have a solution of .5 L of 1.74g of NH3 and 4.15g of (NH4)2SO4. How many ml of 14M HCl is required for a pH of 9.00?
I don't know how to go finish this one. I thought about figuring the concentrations of each (NH3, NH4), making an ice table and solving for the additino of H+. But then i don't know how to get ml, cuz what i solved would be in conc of H+ wouldn't it? :s
(Explanation would be greatful, thanks)
1 answer
1 answer