I don't get this at all.
A beaker with 105mL of an acetic acid buffer with a pH of 5.00 is sitting on a bench top. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
pH = pKa + log(b/a)
5.00 = 4.74 + log b/a
solve for b/a which has two unknowns. The second equation you need is
b + a = 0.1M
Solve the two equations simultaneously for CH3COOH and CH3COO^- and that's the first part of the problem. Post back with this work if you need help finishing with the second part.
need more help
need more help
To determine how the pH will change after adding the HCl solution, we need to first calculate the moles of HCl that are added to the beaker. Then, we can use the Henderson-Hasselbalch equation to calculate the change in pH.
Step 1: Calculate moles of HCl added:
Molarity (M) = moles/volume (L)
Given:
Volume of HCl solution added = 8.40 mL = 8.40/1000 L = 0.0084 L
Molarity of HCl solution = 0.470 M
Moles of HCl added = Molarity × Volume
= 0.470 M × 0.0084 L
= 0.003948 moles
Step 2: Determine the change in moles of acetic acid and conjugate base:
Since the acetic acid buffer is a weak acid and its conjugate base (acetate) is a weak base, they will react with the added HCl.
The balanced equation for the reaction between acetic acid (HA) and HCl is:
HA + HCl → H2O + A^- + Cl^-
In this reaction, 1 mole of HA reacts with 1 mole of HCl to produce 1 mole of A^- (acetate anion).
Therefore, the change in moles of acetic acid and acetate can be calculated as:
Change in moles of acid (HA) = moles of HCl added = 0.003948 moles
Change in moles of conjugate base (A^-) = moles of HCl added = 0.003948 moles
Step 3: Calculate the new concentrations of acid and conjugate base:
The initial concentration of acid and conjugate base in the buffer is 0.100 M. Since the added volumes are negligible compared to the initial volume, the total volume can be assumed to remain approximately constant. Therefore, the change in concentration of acid and conjugate base will be equal to the change in moles.
New concentration of acid (HA) = Initial concentration - Change in concentration
= 0.100 M - 0.003948 moles
≈ 0.096 moles
New concentration of conjugate base (A^-) = Initial concentration + Change in concentration
= 0.100 M + 0.003948 moles
≈ 0.104 moles
Step 4: Apply the Henderson-Hasselbalch equation to calculate the new pH:
The Henderson-Hasselbalch equation is given as:
pH = pKa + log10 ([A^-] / [HA])
Given:
pKa of acetic acid = 4.760
Substituting the values:
pH = 4.760 + log10 (0.104 moles / 0.096 moles)
≈ 4.760 + log10 (1.0833)
Using logarithmic properties, we can simplify:
pH ≈ 4.760 + 0.0351
≈ 4.7951
Therefore, the pH will change by approximately 4.7951.