Question
A buffer solution contains 0.120M acetic acid and 0.150M sodium acetate.
a. How many moles of acetic acid and sodium acetate are present in 50.0 ml of solution?
b. if we add 5.55 mL of 0.092M NaOH to the solution in part (a) how many moles of acetic acid , sodium acetate, and NaOH will be present after the reaction is finished?
c. if we add 0.50 ml of 0.087 M HCI to the solution in part (a), how many moles of acetic acid, sodium acetate, and HCl will be present after the reaction is done?
a. How many moles of acetic acid and sodium acetate are present in 50.0 ml of solution?
b. if we add 5.55 mL of 0.092M NaOH to the solution in part (a) how many moles of acetic acid , sodium acetate, and NaOH will be present after the reaction is finished?
c. if we add 0.50 ml of 0.087 M HCI to the solution in part (a), how many moles of acetic acid, sodium acetate, and HCl will be present after the reaction is done?
Answers
a. M x L = moles.
b. CH3COOH + NaOH ==> CH3COONa + H2O
I...6 mmols....0.......7.5 mmoles
C... 0........0.51 mmols..0
E...6-0.511 ....0.......7.5+0.511
I stands for initial
C stands for change.
E stands for equilibrium.
Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros.
c. done as in b.
b. CH3COOH + NaOH ==> CH3COONa + H2O
I...6 mmols....0.......7.5 mmoles
C... 0........0.51 mmols..0
E...6-0.511 ....0.......7.5+0.511
I stands for initial
C stands for change.
E stands for equilibrium.
Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros.
c. done as in b.
Thank you for your help!!