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
the buffer?
(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? =\
3 answers
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
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
1 answer