A beaker with 195 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.20 mL of a 0.260 it M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Question

A beaker with 195 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.20 mL of a 0.260 it M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.

Answer 1

Call base = B (acetate ion).

Call acid = A (acetic acid).
A + B = 0.1 M
pH = pKa + log (B/A).
Two equations and two unknowns. Solve for A and B.
You know pH and pKa, so solve for B/A.
Then A + B = 0.1
That is the first part of the problem.
For the second part, recognize that
HCl + B^- ==>A + Cl^- (HB is the acid, A).
You know B and you know how much HCl is added, so redo the Henderson-Hasselbalch equation to solve for the new pH.
Post your work if you get stuck.

Answer 2

who knows,but are you good at chemistry, because I need help baddly

Answer 3

To determine how much the pH will change after adding HCl to the acetic acid buffer, you can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:
- pH is the current pH of the buffer solution.
- pKa is the dissociation constant of acetic acid (given as 4.760).
- [A-] and [HA] are the concentrations of the conjugate base and acid, respectively.

First, let's calculate the initial concentrations of [A-] and [HA]:

Total volume of the buffer solution = 195 mL + 8.20 mL = 203.20 mL = 0.20320 L

Molarity of the buffer = 0.100 M

Moles of acid = Molarity × Volume (in liters)
= 0.100 M × 0.20320 L
≈ 0.02032 moles

Since the acetic acid and its conjugate base are in equimolar amounts, each will have a final concentration of:
0.02032 moles / 0.20320 L = 0.100 M

Now, let's calculate the concentrations after adding HCl:

Initial volume of HCl added = 8.20 mL = 0.00820 L
Molarity of HCl added = 0.260 M

Using the dilution formula:

M1V1 = M2V2

(0.100 M)(0.20320 L) = (0.100 M + M2)(0.20320 L + 0.00820 L)

Solving for M2:

M2 = [(0.100 M)(0.20320 L)] / (0.20320 L + 0.00820 L)
≈ 0.0975 M

After mixing, the concentration of the acetic acid and its conjugate base in the buffer will be approximately 0.0975 M.

Now, let's calculate the new pH using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
= 4.760 + log(0.0975 M/0.0975 M)
= 4.760 + log(1)
= 4.760 + 0
= 4.760

Therefore, the pH of the buffer solution will remain unchanged after adding the HCl solution.