0.1M OF NaOH in 5ml is mixed with 0.1M of CH3COOH in 45ml, what is the concentration of CH3COO- in equilibrium mixture?
mols CH3COOH = 0.1 x 0.045L = 0.00450
mols NaOH = 0.1 x 0.005 = 0.0005
.....NaOH + CH3COOH ==> CH3COONa + H2O
I..0.0005....0.0045.......0..........0
C.-0.0005...-0.0005....-0.0005...+0.0005
E....0.......0.0040.....0.0005....0.0005
(CH3COO^-) = mols/L = 0.0005/0.050 = ?
thanks DRBob....how can i find the amount of CH3COOH undissociated in the equilibrium mixture?
To find the concentration of CH3COO- in the equilibrium mixture, we need to determine the amount of CH3COOH and NaOH that react with each other and form CH3COO- ions. This reaction is an acid-base reaction, where NaOH (a strong base) reacts with CH3COOH (a weak acid) to form CH3COO- ions.
First, let's determine the number of moles of NaOH and CH3COOH in the given solutions:
Number of moles of NaOH = concentration (in M) × volume (in L)
= 0.1 M × 0.005 L
= 0.0005 moles
Number of moles of CH3COOH = concentration (in M) × volume (in L)
= 0.1 M × 0.045 L
= 0.0045 moles
Since NaOH and CH3COOH react in a 1:1 ratio, the limiting reactant is CH3COOH since it has fewer moles. This means that all of the CH3COOH will react, and any remaining NaOH will be in excess.
Now, let's calculate how many moles of CH3COO- are formed:
Number of moles of CH3COO- = number of moles of CH3COOH reacted
= 0.0045 moles
To determine the concentration of CH3COO- in the final mixture, we need to consider the total volume of the mixture. The total volume is the sum of the volumes of the NaOH and CH3COOH solutions:
Total volume = volume of NaOH solution + volume of CH3COOH solution
= 0.005 L + 0.045 L
= 0.05 L
Finally, we can calculate the concentration of CH3COO- in the equilibrium mixture:
Concentration of CH3COO- = moles of CH3COO- / total volume
= 0.0045 moles / 0.05 L
= 0.09 M
Therefore, the concentration of CH3COO- in the equilibrium mixture is 0.09 M.