2. A buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate: CH
ID: 957231 • Letter: 2
Question
2. A buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate: CH COOH aq) HO e) CH3COO (aq) H3O (aq) 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 0.092 M NaOH to the solution in part (a), how many moles of acetic acid, sodium acetate, and NaOH will be present after the reaction has finished? (The net ionic equation for the reaction that occurs with NaOH looks like below.) CH3COOH(aq) OH (aq) CH3COO (aq) H2O(e) c. If we add 0.50 mL of 0.087 M HCl to the solution in part (a), how many moles of acetic acid, sodium acetate, and HCl will be present after the reaction is complete? CH COO (aq) H3O (aq) CH COOHOaq) H2O(e) 3. Determine the pH you expect to find for the three solutions in question 2.Explanation / Answer
a.
moles of acetic acid = (0.120 M)*(50.0 mL) = 6.00 mmol = 6.00*10^-3 mol
moles of sodium acetate = (0.150 M)*(50.0 mL) = 7.50 mmol = 7.50*10^-3 mol
b.
moles of NaOH = (0.092 M)*(5.55 mL) = 0.5106 mmol = 0.5106*10^-3 mol
After reaction:
moles of acetic acid = 6.00 mmol - 0.5106 mmol = 5.4894 mmol = 5.4894*10^-3 mol
moles of sodium acetate = 7.50 mmol + 0.5106 mmol = 8.0106 mmol = 8.0106*10^-3 mol
moles of NaOH = 0
c.
moles of HCl = (0.087 M)*(0.50 mL) = 0.0435 mmol = 0.0435*10^-3 mol
After reaction:
moles of acetic acid = 6.00 mmol + 0.0435 mmol = 6.0435 mmol = 6.0435*10^-3 mol
moles of sodium acetate = 7.50 mmol - 0.0435 mmol = 7.4565 mmol = 7.4565*10^-3 mol
moles of HCl = 0
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