A 190 mL sample of 0.293 M NaCH3CO2(aq) is diluted to 750 mL. What is the concentration of the acetic acid at equilibrium? Ka =1.8×10−5.
Answer in units of mol/L.
To find the concentration of acetic acid at equilibrium, we need to consider the dissociation of sodium acetate (NaCH3CO2) in water. NaCH3CO2 dissociates into sodium ions (Na+) and acetate ions (CH3CO2-), and the acetate ions further react with water to form acetic acid (CH3COOH) and hydroxide ions (OH-). This reaction is in equilibrium and can be represented as follows:
CH3CO2- (aq) + H2O ⇌ CH3COOH (aq) + OH- (aq)
Given that the initial concentration of NaCH3CO2 is 0.293 M and the volume is 190 mL, we can calculate the initial number of moles:
moles of NaCH3CO2 = initial concentration × volume
= 0.293 M × 0.190 L
= 0.05567 moles
Next, we dilute the solution to a final volume of 750 mL. Since the amount of solute remains constant during dilution, the number of moles of NaCH3CO2 is still 0.05567 moles.
Now, we can use the equation for the dissociation of NaCH3CO2 to determine the concentration of acetic acid at equilibrium. However, we need to account for the presence of hydroxide ions during the reaction.
The equilibrium constant, Ka, is defined as [CH3COOH] / ([CH3CO2-] × [OH-]). Since Ka = 1.8 × 10^−5, we can rearrange the equation to solve for [CH3COOH]:
[CH3COOH] = Ka × ([CH3CO2-] × [OH-])
Since [CH3CO2-] = 0.05567 moles (initial number of moles) and the concentration of OH- can be determined by using the concentration of NaCH3CO2 and the volume of the solution, we can calculate [CH3COOH]:
[OH-] = ([NaCH3CO2] - [CH3CO2-]) / V
Note: [NaCH3CO2] is the concentration of the initially dissolved NaCH3CO2, which is 0.293 M.
[OH-] = (0.293 M - 0.05567 moles / 0.750 L
= 0.31406 M
Now we can substitute the values into the equation:
[CH3COOH] = Ka × (0.05567 moles × 0.31406 M)
= 1.8 × 10^−5 × (0.05567 moles × 0.31406 M)
= 9.38 × 10^−6 moles / L
Therefore, the concentration of the acetic acid at equilibrium is approximately 9.38 × 10^−6 mol/L.