I understand part (I) but We have only briefly gone over the HH equation in class and I understand how to use it to get the pH but I really don't understand how to use it in part (II), assuming I should be using it in this question?
(I)Calculate the pH of an acetate buffer containing 27.5 mM acetic acid (pKa is 4.74) and 50 mM
sodium acetate.
(II) When a 100 mM aq solution of sodium acetate was prepared in the lab it had a
pH of 8.9. Calculate the amount of acetic acid (in g) to be added to 0.5 L of this sodium acetate
solution in order to bring its pH to the target value of 5.00.
For part II, plug into the HH equation.
5.00 = 4.74 + log (base)/(acid)
5.00 = 4.74 + log (100*0.5)/(x)
Solve for x = mmols acetic acid and convert that to grams acetic acid.
To solve part (II) of the question, you can use the Henderson-Hasselbalch equation again. The equation is:
pH = pKa + log ([A-] / [HA])
Where:
- pH is the pH of the solution (target value)
- pKa is the acid dissociation constant of the weak acid (given as 4.74)
- [A-] is the concentration of the conjugate base (sodium acetate in this case)
- [HA] is the concentration of the weak acid (acetic acid in this case)
In this case, you want to find the amount of acetic acid (in grams) to be added to 0.5 L of the sodium acetate solution to bring its pH to 5.00.
Here's a step-by-step approach to solve the problem:
1. Convert the given pH value of 5.00 to the H+ ion concentration using the formula: [H+] = 10^(-pH).
[H+] = 10^(-5.00) = 1.0 x 10^(-5) M
2. Use the Henderson-Hasselbalch equation to calculate the ratio of [A-] to [HA] needed to achieve the desired pH:
5.00 = 4.74 + log ([A-] / [HA])
3. Rearrange the equation to solve for the ratio [A-] / [HA]:
log ([A-] / [HA]) = 5.00 - 4.74
log ([A-] / [HA]) = 0.26
4. Take the antilog of both sides to get rid of the logarithm:
[A-] / [HA] = 10^(0.26)
[A-] / [HA] = 1.77
5. Since you have the concentration of sodium acetate (50 mM), you can set up the equation:
1.77 = [50 mM] / [HA]
6. Rearrange the equation to solve for [HA]:
[HA] = [50 mM] / 1.77
[HA] = 28.25 mM
7. Convert the concentration of acetic acid ([HA]) from millimoles per liter (mM) to grams per liter (g/L):
Molecular weight of acetic acid (CH3COOH) = 60.05 g/mol
[HA] in g/L = [HA] in mM x molecular weight of acetic acid / 1000
= 28.25 mM x 60.05 g/mol / 1000
= 1.696 g/L
8. Finally, calculate the amount of acetic acid in grams required for 0.5 L of the sodium acetate solution:
Amount of acetic acid (in g) = [HA] in g/L x volume of solution (in L)
= 1.696 g/L x 0.5 L
= 0.848 g
Therefore, you need to add approximately 0.848 grams of acetic acid to 0.5 L of the sodium acetate solution to bring its pH to the target value of 5.00.