We will assume that the volumes are additive.
mols NaOH initially = M x L = 0.124 x 0.500 = 0.062
mols CH3COOH initially = 0.100 x 0.500 = 0.05
The ICE chart looks like this.
........NaOH + CH3OOH ==> CH3COONa + H2O
I.......0.062........0.05...................0...................0
C.........-0.05.....-0.05.................0.05...........0.05
E.......0.012...........0...................0.05..........0.05
From the E line, (Na^+) = 0.012 from NaOH left + 0.05 from CH3COOH formed = 0.062 mols and that is in 1 L assume the volumes are additive so (Na^+) = mols/L = ?
From the E line, (CH3COO^-) = 0.05 mols/L = ?
From the E line, and ignoring any hydrolysis of the acetate ion,
(CH3COOH) = 0
(H^+) is calculated from
(H^+)(OH^-) = Kw = 1E-14.
The (OH^-) from the E line is 0.012/1L = ? and plug that into the Kw expression and solve for H^+.
Check this thoroughly.
A solution is made by mixing exactly 500 mL of 0.124 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
Enter your answer in scientific notation.
[OH−]
[CH3COOH]
× 10
Enter your answer in scientific notation.
[Na+]
[CH3COO−]
1 answer
2 answers