Let's call CH3COOH simply HAc. Saves typing that way.
.
millimols HAc initially = mL x M = 500 x 0.1 = 50
millimols NaOH initially = 500 x 0.156 = 78
........HAc + NaOH --> NaAc + H2O
I.........50..........78...........0...........0
C......-50..........-50.........+50.......50
E.........0...........28..........50..........50
Concentration of the species is millimols/mL. Remember mL is now 1000 because 500 mL has been added to 500 mL to make 1000 mL.
(HAc) = 0
(OH^-) = 28/1000 = ?
(H^+)(OH^-) = Kw. You know Kw and OH, solve for H.
(Na^+) is from the NaOH
(Ac^-) is from the NaAc
Etc
Post your work if you get stuck.
A solution is made by mixing exactly 500 mL of 0.156 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below. Ka of CH3COOH is 1.8 × 10−5
[H+]
× 10 M
Enter your answer in scientific notation.
[OH−]
M
[CH3COOH]
× 10 M
Enter your answer in scientific notation.
[Na+]
M
[CH3COO−]
M
1 answer