What is the new concentration of acetic acid after 120mL of 0.201M acetic acid is mixed with 160mL of 0.248M sodium acetate
Im not sure how to start this one would it be (120x.201)+(160x.248)
This makes a buffered solution; however, the problem doesn't ask for pH. I think this is nothing more than a dilution problem.
new (HAc) = 0.201 x [120/(120+160) ]= ?
thank you
To solve this problem, you need to apply the concept of dilution.
First, let's calculate the moles of acetic acid (CH3COOH) and sodium acetate (CH3COONa) in each solution.
For the 120 mL of 0.201 M acetic acid solution:
Moles of acetic acid = concentration (M) × volume (L)
Moles of acetic acid = 0.201 M × 0.120 L = 0.02412 moles
For the 160 mL of 0.248 M sodium acetate solution:
Moles of sodium acetate = concentration (M) × volume (L)
Moles of sodium acetate = 0.248 M × 0.160 L = 0.03968 moles
Next, let's determine which compound is the limiting reagent and how much of it remains after the reaction.
The balanced equation for the reaction between acetic acid and sodium acetate is:
CH3COOH + CH3COONa → CH3COONa + CH3COOH
From the equation, we can see that 0.02412 moles of acetic acid reacts with an equal amount of sodium acetate. Therefore, acetic acid is the limiting reagent.
Since all the acetic acid (0.02412 moles) reacts, there is no acetic acid left in the final solution.
Finally, let's calculate the concentration of acetic acid in the final solution.
To find the total volume of the final solution, add the volumes of the two solutions:
Final volume = 120 mL + 160 mL = 280 mL = 0.280 L
Since all the acetic acid reacts, the final concentration is zero.