You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic
ID: 557208 • Letter: Y
Question
You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic acid and 0.100 M sodium acetate solutions. Calculate the volume of 0.100 M acetic acid required to prepare 60.0 mL of a buffer of pH 4.5. 38.2 mL 1 pts You are correct. Previous Tries Your receipt no. is 168-1656 Calculate the volume of 0.100 M sodium acetate required to prepare 60.0 mL of a buffer of pH 4.5. 21.8 mL 1 pts You are correct. Previous Tries Your receipt no. is 168-250 Calculate the concentration [HA] in this buffer. 0.0637 M 1 pts You are correct. Previous Tries Your receipt no. is 168-1166 Calculate the concentration [A 1 in this buffer. 0.0363 M 1 ptsExplanation / Answer
The buffer contains 38.2 mL of 0.0637 M HA and 21.8 mL of 0.0363 M A-. Find out the number of millimoles of HA and A- in the buffer.
HA = (38.2 mL)*(0.0637 M) = 2.43334 mmole.
A- = (21.8 mL)*(0.0363 M) = 0.79134 mmole.
12.7 mL of 0.1000 M NaOH has been added to the buffer; the millimoles of NaOH added = (12.7 mL)*(0.1000 M) = 1.27 mmole.
HA reacts with NaOH as below.
HA (aq) + NaOH (aq) --------> NaA (aq) + H2O (l)
NaA furnishes A-; as per the stoichiometry of the reaction,
1 mole HA =1 mole NaOH = 1 mole A-.
Therefore, 1.27 mmole A- is formed at equilibrium and 1.27 mmole HA is neutralized.
Therefore, millimoles of A- at equilibrium = (0.79134 + 1.27) mmole = 2.06134 mmole.
Millimoles of HA at equilibrium = (2.43334 – 1.27) mmole = 1.16334 mmole.
The volume of the solution stays constant; hence, we can write, [A-]/[HA] = (mmoles of A-)/(mmoles of HA) = (2.06134 mmole)/(1.16334 mmole) = 1.771915.
The pKa of acetic acid, HA is 4.74; use the Henderson-Hasslebahc equation.
pH = pKa + log [A-]/[HA] = 4.74 + log (1.771915) = 4.988443 4.99 (ans).
7.3 mL of 0.1000 M HCl has been added to the buffer; the millimoles of HCl added = (7.3 mL)*(0.1000 M) = 0.73 mmole.
A- reacts with HCl as below.
A- (aq) + HCl (aq) --------> HA (aq) + Cl- (l)
1 mole A- =1 mole HCl = 1 mole HA.
Therefore, 0.73 mmole A- reacts with 0.73 mmole HCl to form 0.73 mmole HA.
Therefore, millimoles of A- at equilibrium = (0.79134 – 0.73) mmole = 0.06134 mmole.
Millimoles of HA at equilibrium = (2.43334 + 0.73) mmole = 3.16334 mmole.
The volume of the solution stays constant; hence, we can write, [A-]/[HA] = (mmoles of A-)/(mmoles of HA) = (0.06134 mmole)/(3.16334 mmole) = 0.019391.
The pKa of acetic acid, HA is 4.74; use the Henderson-Hasslebahc equation.
pH = pKa + log [A-]/[HA] = 4.74 + log (0.019391) = 3.027600 3.03 (ans).
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