Hi, here the question
Calculate the mass of sodium acetate formed when 5.0g of acetic acid is added to 100mL of 0.50 mol L^-1 sodium hydroxide solution. Show all solution.
I'm not really sure where to start on this one. I know that 5.0g of acetic acid would be converted to moles and then added to solution to work out the new molarity. No idea though from there!
Please help! Thanks :)
Calculate the mass of sodium acetate formed when 5.0g of acetic acid is added to 100mL of 0.50 mol L^-1 sodium hydroxide solution. Show all solution.
I'm not really sure where to start on this one. I know that 5.0g of acetic acid would be converted to moles and then added to solution to work out the new molarity. No idea though from there!
Please help! Thanks :)
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the chemical equation for this reaction would be CH3COOH + NaOH ---> CH3COONa + H2O
We have to work out the excess reactant, so we convert both to mole and see which one has more.
n(CH3COOH) =5g/60g per mol = 0.0833mol
n(NaOH) = 0.5Mx0.1L = 0.05mol
So, CH3COOH is in excess. From this we can tell we need to use the values for NaOH to get the answer.
Since the number of mole of NaOH is 0.05 mol then the Sodium Acetate would also have 0.05mol as there is a 1:1 ratio between the two.
from here we can work out the mass of Sodium Acetate.
m(CH3COONa) = nM = 0.05x82= 4.1g
And that's your answer (: Hope it helped!
We have to work out the excess reactant, so we convert both to mole and see which one has more.
n(CH3COOH) =5g/60g per mol = 0.0833mol
n(NaOH) = 0.5Mx0.1L = 0.05mol
So, CH3COOH is in excess. From this we can tell we need to use the values for NaOH to get the answer.
Since the number of mole of NaOH is 0.05 mol then the Sodium Acetate would also have 0.05mol as there is a 1:1 ratio between the two.
from here we can work out the mass of Sodium Acetate.
m(CH3COONa) = nM = 0.05x82= 4.1g
And that's your answer (: Hope it helped!