What is the new pH when 1mL of 1M HCl os added to 5mL of 0.1M Phosphate buffer (pH 7.4)?
I assume you are using pK2 for pKa. You need to look up the value in your text or notes. My OLD text lists pK2 as 7.21.
First,
7.40 = 7.21 + log (base/acid)
Solve for B/A = 1.549 which makes
B = 1.549*A
Then B + A = 0.1, solve these two equations simultaneously and I arrive at
A = 0.03923M and B = 0.06077M
5 mL of the buffer gives us
0.03923 x 5 mL = 0.196 mmoles A
0.06077 x 5 mL = 0.304 mmoles B.
...............Base + H^+ ==> Acid
initial mmols..0.304..0.......0.196
change mmols...-1.....1.........1
final mmols....0.204..0........1.196
Substitute into the HH equation. I obtain something like 6.3 or so but you need to confirm that. Check my work.
To determine the new pH when 1mL of 1M HCl is added to 5mL of 0.1M Phosphate buffer (pH 7.4), you can use the Henderson-Hasselbalch equation. This equation is commonly used to calculate the pH of a buffer solution.
The Henderson-Hasselbalch equation is defined as:
pH = pKa + log ([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the logarithmic acid dissociation constant of the weak acid
[A-] = the concentration of the conjugate base
[HA] = the concentration of the weak acid
In this case, the phosphate buffer acts as a weak acid, and its pKa can be assumed to be around 7.4 due to the pH of the buffer solution. The concentration of the conjugate base ([A-]) can be calculated by diluting the initial concentration of the buffer (0.1M phosphate) with the total volume of the solution (5mL + 1mL).
Using the dilution equation:
C1V1 = C2V2
Where:
C1 = initial concentration of the buffer (0.1M)
V1 = initial volume of the buffer (5mL)
C2 = final concentration of the buffer-conjugate base mixture
V2 = final volume of the buffer-conjugate base mixture (5mL + 1mL)
Solving for C2:
C2 = (C1 * V1) / V2
C2 = (0.1M * 5mL) / 6mL
C2 = 0.0833M
Now that we know the concentration of the conjugate base ([A-]) is 0.0833M, we can substitute it into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = 7.4 + log (0.0833M/0.1M)
Now, solve for pH:
pH = 7.4 + log (0.833)
pH = 7.4 - 0.08
pH ≈ 7.32
Therefore, when 1mL of 1M HCl is added to 5mL of 0.1M Phosphate buffer (pH 7.4), the new pH of the solution is approximately 7.32.