posted by Chelsea .
If you add 5.0 mL of 0.50 M NaOH solution to 20.0 mL to Buffer C, what is the change in pH of
(where buffer C is 8.203 g sodium acetate with 100.0 mL of 1.0 M acetic acid)
I have calculated the pH of buffer C to be 4.74.
Now what? =\
I am a retard, I have figured this out
couple someone please still explain? and explain how the 4.74 pH was found? thanks
could someone please still explain? and explain how the 4.74 pH was found? thanks
pH = pKa + log ([A-]/[HA])
We ned to determine the concentration of the sodium acetate:
Molar mass CH3COONa = 82.0340 g/mol
Therefore you are adding 1mol CH3COOH per litre = 1.00M
pKa CH3COOH = 4.75
pH = 4.75 + log (1/1)
pH = 4.75 as you got - OK.
Question 2: If you add 5 ml of 0.50M NaOH solution to 20ml of the buffer, what is the change in pH?
The answer should be that the pH does not change very much - that is what a buffer solution is all about. I am sure that there is a simple way to make this calculation - but I do not know it. So let us try and do it from first principles:
You have a buffer solution that is 1.00M in CH3COOH and 1.00M in CH3COONa.
If you add some NaOH you react with the CH3COOH , reducing its concentration and increasing the concentration of CH3COONa
Mol CH3COOH in 20 mL of 1.00M = 20/1000*1 = 0.02mol CH3COOH
Mol CH3COONa in 20mL of 1.00M = 0.02mol CH2COONa
Mol NaOH in 5.0mL of 0.50M = 5/1000*0.5 = 0.0025mol NaOH
The NaOH reacts with the CH3COOH to form 0.0025 mol CH3COONa and 0.0025 mol CH3COOH is removed
You end up with
CH3COOH = 0.02mol - 0.0025 mol = 0.0175mol CH3COOH in 25mL total solution:
Molarity of CH3COOH = 0.0175/25*1000 = 0.70M CH3COOH
CH3COONa = 0.02mol + 0.0025mol = 0.0225 mol CH3COONa in 25mL total solution
Molarity of CH3COONa = 0.0225/25*1000 = 0.9M CH3COONa
pH = pKa + log([A-]/[HA])
pH = 4.75 + log( 0.9/0.7)
pH = 4.75 + log 1.286
pH = 4.75 + 0.11
pH = 4.86