100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 15.2mL of 0.35M NaOH. What is the pH of the resulting solution?
100 mL of buffer solution that is 0.15M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 28.3mL of 0.25M HCl. What is the pH of the resulting solution?
NaOH will react with HA.
Calculate moles, then:
0.00532 mole of NaOH+ 0.020 NaA = x : new amount of NaA
0.0150 HA-0.00532 moles of NaOH=y: new amount of HA
-log(KA)=pka
pH=pka+log[x/y]
Second problem is essentially the same:
HCl is added to the amount of HA and subtracted from the amount of NaA.
To find the pH of the resulting solution, we need to consider the reaction between the components of the buffer solution (HA and NaA) and the added acid/base (NaOH or HCl). We'll use the Henderson-Hasselbalch equation to calculate the pH.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
First, let's solve the first question.
1. Calculate the moles of HA and NaA in the initial buffer solution:
moles of HA = 0.15 M * 0.100 L = 0.015 mol
moles of NaA = 0.20 M * 0.100 L = 0.020 mol
2. Calculate the moles of OH- ions added from NaOH:
moles of NaOH = 0.35 M * 0.0152 L = 0.00532 mol (since NaOH is a strong base, it dissociates fully into Na+ and OH- ions)
moles of OH- in solution after addition = 0.00532 mol
3. Calculate the remaining moles of HA and A- (formed by the reaction of HA with OH-):
moles of HA remaining = moles of HA initial - moles of OH- ions added = 0.015 mol - 0.00532 mol = 0.00968 mol
moles of A- formed = moles of OH- ions added = 0.00532 mol
4. Calculate the concentrations of HA and A- in the resulting solution:
[H+] = [A-] = moles/volume = 0.00968 mol / (0.100 L + 0.0152 L) = 0.085 M
5. Calculate pKa using the given Ka value:
Ka = 6.8x10^-5 = [H+][A-]/[HA]
Since [H+] = [A-] = 0.085 M,
6.8x10^-5 = (0.085)(0.085)/[HA]
[HA] = (0.085)(0.085)/6.8x10^-5 = 0.106 M
6. Substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]) = -log(6.8x10^-5) + log(0.085/0.106)
pH = -(-4.17) + (-0.087) = 4.17 + 0.087 = 4.257
Therefore, the pH of the resulting solution is approximately 4.26.
Now let's solve the second question.
The process is similar, but instead of OH- ions from NaOH, we have H+ ions from HCl.
1. Calculate the moles of HA and NaA in the initial buffer solution:
moles of HA = 0.15 M * 0.100 L = 0.015 mol
moles of NaA = 0.20 M * 0.100 L = 0.020 mol
2. Calculate the moles of H+ ions added from HCl:
moles of HCl = 0.25 M * 0.0283 L = 0.007075 mol
moles of H+ ions in solution after addition = 0.007075 mol
3. Calculate the remaining moles of HA and A- (formed by the reaction of HA with H+):
moles of HA remaining = moles of HA initial - moles of H+ ions added = 0.015 mol - 0.007075 mol = 0.007925 mol
moles of A- formed = moles of H+ ions added = 0.007075 mol
4. Calculate the concentrations of HA and A- in the resulting solution:
[H+] = [A-] = moles/volume = 0.007925 mol / (0.100 L + 0.0283 L) = 0.0526 M
5. Calculate pKa using the given Ka value:
Ka = 6.8x10^-5 = [H+][A-]/[HA]
Since [H+] = [A-] = 0.0526 M,
6.8x10^-5 = (0.0526)(0.0526)/[HA]
[HA] = (0.0526)(0.0526)/6.8x10^-5 = 0.04056 M
6. Substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]) = -log(6.8x10^-5) + log(0.0526/0.04056)
pH = -(-4.17) + (0.222) = 4.17 + 0.222 = 4.392
Therefore, the pH of the resulting solution is approximately 4.39.