A buffer solution contains 0.120M acetic acid and 0.150M sodium acetate.
a. How many moles of acetic acid and sodium acetate are present in 50.0 ml of solution?
b. if we add 5.55 mL of 0.092M NaOH to the solution in part (a) how many moles of acetic acid , sodium acetate, and NaOH will be present after the reaction is finished?
c. if we add 0.50 ml of 0.087 M HCI to the solution in part (a), how many moles of acetic acid, sodium acetate, and HCl will be present after the reaction is done?
a. M x L = moles.
b. CH3COOH + NaOH ==> CH3COONa + H2O
I...6 mmols....0.......7.5 mmoles
C... 0........0.51 mmols..0
E...6-0.511 ....0.......7.5+0.511
I stands for initial
C stands for change.
E stands for equilibrium.
Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros.
c. done as in b.
Thank you for your help!!
a. To find the moles of acetic acid and sodium acetate in the solution, we will use the formula:
Moles = concentration (M) × volume (L)
Given:
Concentration of acetic acid = 0.120 M
Concentration of sodium acetate = 0.150 M
Volume of solution = 50.0 mL = 0.050 L
Moles of acetic acid = 0.120 M × 0.050 L = 0.006 moles
Moles of sodium acetate = 0.150 M × 0.050 L = 0.0075 moles
b. If we add 5.55 mL of 0.092 M NaOH to the solution in part (a), we can calculate the moles of acetic acid, sodium acetate, and NaOH after the reaction is complete.
Given:
Volume of NaOH added = 5.55 mL = 0.00555 L
Concentration of NaOH = 0.092 M
NaOH reacts with acetic acid, forming water and sodium acetate. The reaction equation is:
NaOH + CH3COOH -> CH3COONa + H2O
From the reaction equation, we can see that the mole ratio between NaOH and acetic acid is 1:1. Therefore, the moles of acetic acid consumed will be equal to the moles of NaOH added.
Moles of acetic acid after the reaction = 0.006 moles - 0.00555 moles = 0.00045 moles
Moles of sodium acetate after the reaction will remain the same as before, 0.0075 moles.
Moles of NaOH after the reaction = 0.092 M × 0.00555 L = 0.0005094 moles
c. If we add 0.50 mL of 0.087 M HCl to the solution in part (a), we can calculate the moles of acetic acid, sodium acetate, and HCl after the reaction is complete.
Given:
Volume of HCl added = 0.50 mL = 0.00050 L
Concentration of HCl = 0.087 M
HCl reacts with sodium acetate, forming acetic acid and sodium chloride. The reaction equation is:
HCl + CH3COONa -> CH3COOH + NaCl
From the reaction equation, we can see that the mole ratio between HCl and sodium acetate is 1:1. Therefore, the moles of sodium acetate consumed will be equal to the moles of HCl added.
Moles of acetic acid after the reaction will remain the same as before, 0.006 moles.
Moles of sodium acetate after the reaction = 0.0075 moles - 0.00050 moles = 0.007 moles
Moles of HCl after the reaction = 0.087 M × 0.00050 L = 0.0000435 moles
To answer these questions, we need to use the molarity-concentration relationship and the volume of the solution. Here's how you can calculate the number of moles:
a. For acetic acid:
Molarity of acetic acid = 0.120 M
Volume of solution = 50.0 mL = 0.0500 L
Number of moles of acetic acid = Molarity x Volume
= 0.120 M x 0.0500 L
= 0.00600 mol
For sodium acetate:
Molarity of sodium acetate = 0.150 M
Volume of solution = 50.0 mL = 0.0500 L
Number of moles of sodium acetate = Molarity x Volume
= 0.150 M x 0.0500 L
= 0.00750 mol
b. When NaOH reacts with acetic acid, it forms sodium acetate and water. So, we need to consider the reaction:
CH3COOH + NaOH -> CH3COONa + H2O
Given:
Volume of NaOH = 5.55 mL = 0.00555 L
Molarity of NaOH = 0.092 M
Before the reaction, the number of moles of acetic acid and sodium acetate remain the same as in part (a). Let's calculate the remaining moles after the reaction:
Moles of acetic acid = 0.00600 mol (from part a)
Moles of sodium acetate = 0.00750 mol (from part a)
Since the ratio of acetic acid to NaOH is 1:1, the moles of NaOH used in the reaction will be the same as the moles of acetic acid:
Moles of NaOH used = Moles of acetic acid = 0.00600 mol
Now, let's calculate the moles of NaOH remaining after the reaction:
Moles of NaOH remaining = Moles of NaOH initially - Moles of NaOH used
= (0.092 M x 0.00555 L) - 0.00600 mol
= 0.0005106 mol
So, the final moles of acetic acid, sodium acetate, and NaOH will be:
Moles of acetic acid = 0.00600 mol (remains the same)
Moles of sodium acetate = 0.00750 mol (remains the same)
Moles of NaOH = 0.0005106 mol (remaining after the reaction)
c. When HCl reacts with sodium acetate, it forms acetic acid and sodium chloride. So, we need to consider the reaction:
CH3COONa + HCl -> CH3COOH + NaCl
Given:
Volume of HCl = 0.50 mL = 0.00050 L
Molarity of HCl = 0.087 M
Before the reaction, the number of moles of acetic acid and sodium acetate remain the same as in part (a). Let's calculate the remaining moles after the reaction:
Moles of acetic acid = 0.00600 mol (from part a)
Moles of sodium acetate = 0.00750 mol (from part a)
Since the ratio of sodium acetate to HCl is 1:1, the moles of HCl used in the reaction will be the same as the moles of sodium acetate:
Moles of HCl used = Moles of sodium acetate = 0.00750 mol
Now, let's calculate the moles of HCl remaining after the reaction:
Moles of HCl remaining = Moles of HCl initially - Moles of HCl used
= (0.087 M x 0.00050 L) - 0.00750 mol
= 0.00004275 mol
So, the final moles of acetic acid, sodium acetate, and HCl will be:
Moles of acetic acid = 0.00600 mol (remains the same)
Moles of sodium acetate = 0.000 mol (fully reacted)
Moles of HCl = 0.00004275 mol (remaining after the reaction)
Note: In the case of part (c), the sodium acetate gets consumed completely in the reaction and is converted into acetic acid, resulting in a decrease in its moles.