A buffer is prepared by adding 0.97 L of 1.1 M HCl to 805 mL of 1.7 M NaHCOO. What is the pH of this buffer? [Ka(HCOOH) = 1.7 × 10-4]
Use the Henderson-Hasselbalch equation. Post your work if you get stuck.
What I did the first time:
pH = -log(1.7*10^-4) + log (1.7/1.1)
I tried it this way before, but it said I got the wrong answer. Does it have something to do with the Ka value?
No. You substituted molarity of the individual solutions BEFORE they were mixed. After mixing the molarity has changed.
initial CH3COO^- = 1.7M x 0.805L =1.3685 moles.
initial HCl = 1.1M x 0.97 L = 1.067 moles.
............CH3COO^- + H^+ ==> CH3COOH
initial mols.1.3685...1.067.....0
change.....-1.067.....-1.067.....1.067
final.......0.3015.....0........1.067
final (CH3COO^-) = 0.3015/total volume.
final (CH3COOH) = 1.067/total volume.
Those are the numbers you substitute into the HH equation.
total volume = 1.775 L
final concentration (CH3COO)= .169859
final concentration (CH3COOH) = .60112676
pH= -log(1.7*10^-4) + log(.169859/.60112676)
= 3.22
is this correct?
1.775 is right.
Concn CH3COO is right.
Concn CH3COOH is right.
Your answer of 3.22 is correct for the value of Ka you show in the problem of 1.7 x 10^-4; however, you may want to check your post and the problem from which you copied it because the Ka for CH3COOH (acetic acid) is 1.7 x 10^-5 (not 10^-4) so the correct answer for your post is 3.22 but the correct answer for an acetic acid/acetate buffer in that proportion is 4.22.
ok thanks!
To find the pH of the buffer solution, we need to determine the concentration of the acid (HCOOH) and its conjugate base (HCOO-) in the solution, and then use the Henderson-Hasselbalch equation.
First, let's calculate the moles of HCl and NaHCOO used:
Moles of HCl = volume (in L) × concentration (in M) = 0.97 L × 1.1 M = 1.067 mol
Moles of NaHCOO = volume (in L) × concentration (in M) = 0.805 L × 1.7 M = 1.3695 mol
The acid (HCOOH) and base (HCOO-) react in a 1:1 ratio, so the moles of HCOO- in the solution will be equal to the moles of NaHCOO used.
Now, let's calculate the final volume of the buffer solution:
Final volume = volume of HCl + volume of NaHCOO = 0.97 L + 0.805 L = 1.775 L
Next, we can determine the concentration of the acid (HCOOH) and its conjugate base (HCOO-) in the final solution:
Concentration of HCOOH = moles of HCOOH / final volume = 1.067 mol / 1.775 L ≈ 0.6 M
Concentration of HCOO- = moles of HCOO- / final volume = 1.3695 mol / 1.775 L ≈ 0.77 M
Now, we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log10(concentration of HCOO- / concentration of HCOOH)
Where pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka). In this case, pKa = -log10(1.7 × 10^(-4)) = 3.77
pH = 3.77 + log10(0.77 M / 0.6 M) ≈ 3.77 + log10(1.283)
To evaluate log10(1.283), use a scientific calculator or logarithm table to find the value. The result will be the final pH of the buffer solution.
Note: The result provided here is an approximation.