I was wondering if you would be able to help me with a Chemistry question about the theoretical pH of a buffer solution. I have to calculate the theoretical pH values expected for a 200mL buffer solution containing a 1:1 ratio of acetic acid and sodium acetate (their concentrations are 0.05M), following the addition of;
a) 10mL of 0.2M HCl and
b) 10mL of 0.2M NaOH.
Attempt for a:
[HCl] = 0.2M V = 10mL/0.01L thus n = 0.002
0.002 x 0.2L = 0.01M
........H3O + A <=> HA + H2O
I = x 0.05. 0.05
C = - 0.01. -0.01 +0.01
E = x-0.01. 0.04. 0.06
pH = 4.74 + log (0.06/0.04)
= 4.926
Although in our experiments the actual pH was 4.17 after 10mL of HCl was added and the pH after 10 mL of NaOH was 4.98
ATTEMPT for b
[NaOH] = 0.2M V = 10mL/0.01L thus n = 0.002
0.002 x 0.2L = 0.01M
-------HA + OH- <=> H2O + A-
I= 0.05. x 0.05
C= -0.01. x-0.01. + 0.01
E = 0.04. 0.06
pH = 4.74 + log(0.04/0.06)
pH = 3.74
So I feel like I have done these the wrong way because it makes more sense but I don't know why or how.
pKa's of phosphoric acid are 2.3, 7.21 and 12.35. If required pH is 6, then, 7.21 will be used. This means, monopotassium dihydrogen phosphate and dipotassium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO4--) potassium salts) will be used.
pH = pKa + log ([A-]/[HA])
Now, for A- you put K2HPO4 concentration, and for HA you put KH2PO4 concentration:
6 = 7.21 + log(A/HA)
log(A/HA)=-1.2
A/HA = 0.063
So, now you know the fold difference between these salts, and the Molarity of the solution would be given to you, and you can calculate the required amount:
HA+A=0.05
A/HA=0.063
HA=0.047M
A=0.003M
This means, you need to put 0.047 moles of KH2PO4 and 0.003 moles of K2HPO4 salts into 1 liters of solution.
for molecular weights: KH2PO4= 39+97=136amu; K2HPO4= 39*2+96=174amu
Thus, 0.047*136=6.392g of KH2PO4 and 0.003*174= 0.522g of K2HPO4 should be added.
But, you should also note that, if |pH-pKa| >1 , then buffer capacity of solution decreases. In this case, it is equal to 1.2
I hope the calculations are clear.
Wish you luck.
I'm a bit confused with your calculations. We were only given one Ka which was for acetate, which I calculated the pKa and used it in the Henderson equation. I don't know where the KH2PO4 is coming from. I'm only meant to calculate it with the information given with this equilibrium
CH3COOH + H2O <---> CH3COONa + H3O
And my teacher said this should be solvable with the ICE analysis table easily. Just a bit confused can you please clear this up? Thank you
To calculate the theoretical pH of a buffer solution, you need to consider the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the pKa (acid dissociation constant) of the weak acid and the ratio of the concentrations of the weak acid and its conjugate base.
First, let me explain the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
where pH is the logarithmic measure of the acidity or basicity of a solution, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
For your buffer solution:
a) 10mL of 0.2M HCl
To calculate the change in concentrations, you need to consider the stoichiometry of the reaction. When 10mL of 0.2M HCl is added to the buffer solution, it reacts with the acetic acid (HA) to form water and acetate ions (A-):
HCl + HA -> H2O + A-
From the balanced equation, you can see that for every molecule of HCl added, one molecule of HA is consumed and one molecule of A- is produced.
Now, calculate the concentrations of HA and A- before and after the reaction:
Initial concentration of HA = 0.05M (since the buffer solution contains a 1:1 ratio of acetic acid and sodium acetate)
Initial concentration of A- = 0.05M (since the buffer solution contains a 1:1 ratio of acetic acid and sodium acetate)
Change in concentration of HA = -0.01M (since 0.01 moles of HA are consumed)
Change in concentration of A- = +0.01M (since 0.01 moles of A- are produced)
Substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = pKa + log (0.01/0.05)
The pKa value for acetic acid is 4.74 (you don't need to calculate this, it's a known value in the case of acetic acid).
pH = 4.74 + log (0.01/0.05)
pH = 4.74 + log (0.2)
pH ≈ 4.74 + (-0.699)
pH ≈ 4.04
So the theoretical pH after adding 10mL of 0.2M HCl is approximately 4.04, which is close to the experimental value of 4.17.
b) 10mL of 0.2M NaOH
To calculate the change in concentrations, you need to consider the stoichiometry of the reaction. When 10mL of 0.2M NaOH is added to the buffer solution, it reacts with the acetate ions (A-) to form water and acetic acid (HA):
NAOH + A- -> H2O + HA
From the balanced equation, you can see that for every molecule of NaOH added, one molecule of A- is consumed, and one molecule of HA is produced.
Now, calculate the concentrations of HA and A- before and after the reaction:
Initial concentration of HA = 0.05M (since the buffer solution contains a 1:1 ratio of acetic acid and sodium acetate)
Initial concentration of A- = 0.05M (since the buffer solution contains a 1:1 ratio of acetic acid and sodium acetate)
Change in concentration of HA = +0.01M (since 0.01 moles of HA are produced)
Change in concentration of A- = -0.01M (since 0.01 moles of A- are consumed)
Substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = pKa + log (0.04/0.06)
Using the pKa value of 4.74:
pH = 4.74 + log (0.04/0.06)
pH = 4.74 + log (0.67)
pH ≈ 4.74 + (-0.176)
pH ≈ 4.56
So the theoretical pH after adding 10mL of 0.2M NaOH is approximately 4.56, which is close to the experimental value of 4.98.
In both cases, the experimental values may differ slightly from the theoretical values due to experimental errors or other factors. But overall, your calculations appear to be correct.