A beaker with 200 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 6.40 mL of a 0.400 \it M \rm HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
I've tried using the Henderson Hasselbalch equation but the answer isn't coming out right.
Perhaps I can help if you will decipher how much and what molarity HCl was added. \it M \rm doesn't make any sense to me.
It's supposed to be 0.400 M HCl.
Can you show me what you've done? You say you don't get the right answer. Do you nave the answer?
pH = pKa + log(B/A) where B = base and A = acid.
5.00 = 4.76 + log(B/A)
Solve for B/A. There are two unknowns here.
Then the problem states that
A + B = 0.1M. A second equation with the same two unknowns. Solve the two equations simultaneously for (A) and (B).
Then millimoles A = 200 mL x (A) = ??
mmoles B = 200 x B = ??
.........base + HCl ==> acid
You add 6.40 mL x 0.400 M HCl = xx mmoles.
Then the acid is increased by xx mmoles and the base is decreased by xx mmoles. Then plug these numbers back into the HH equation to find final pH. My answer is 4.77.
To determine how much the pH will change after adding the HCl solution, we can use the Henderson-Hasselbalch equation. However, there seems to be an error in your calculation. Let's break down the steps to solve the problem correctly.
Step 1: Calculate the initial moles of acetic acid and conjugate base in the buffer solution.
To do this, we need the initial concentrations of acetic acid (CH3COOH) and its conjugate base (CH3COO-) in the buffer. The total molarity given is 0.100 M, so each component has a concentration of 0.050 M (since the buffer contains equal parts of the acid and conjugate base).
Step 2: Determine the number of moles of HCl added.
We are told that the student added 6.4 mL of a 0.400 M HCl solution. To find the number of moles, we multiply the volume by the concentration, then convert to liters:
moles of HCl = (6.4 mL) * (0.400 mol/L) * (1 L / 1000 mL) = 0.00256 mol
Step 3: Calculate the change in concentration for each component.
Since the reaction is between HCl and the conjugate base (CH3COO-), the stoichiometry is 1:1. This means that for every mole of HCl added, an equal amount of the conjugate base will react. Therefore, the change in concentration of the conjugate base is also 0.00256 mol.
Step 4: Calculate the new concentrations of acetic acid and conjugate base.
The initial concentration of acetic acid (CH3COOH) is 0.050 M, and the change is zero.
The initial concentration of the conjugate base (CH3COO-) is 0.050 M, and the change is +0.00256 mol. Therefore, the new concentration of the conjugate base is 0.050 M + 0.00256 mol.
Step 5: Calculate the new pH using the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([conjugate base] / [acid])
pKa = 4.76 (given)
Using the new concentrations obtained in Step 4, plug them into the equation:
pH = 4.76 + log(0.05256 mol / 0.05 mol)
Calculate the result using a logarithm calculator or logarithm rules, and you will obtain the new pH.
By following these steps correctly, you should be able to find the correct change in pH after adding the HCl solution to the buffer.