Calculate the amount of sodium acetate to be added to 100ml of a 0.100M acetic acid (pka=4.76) solution to obtain a buffer of pH 4.50.
pH = pKa + log (base)/(acid)
450 = 4.76 + log (NaAc)/(0.1)
Solve for (NaAc) in molarity.
Then M NaAc = mols NaAc/L NaAc.
You know L and M, solve for mols.
Finally, mols = grams/molar mass. You know molar mass and mols, solve for grams.
I managed to find the mols however the molar mass is not given in the question. How am i supposed to solve for grams? Unless i have to google it which is 82.03 g/mol. correct?
To calculate the amount of sodium acetate needed to create a buffer of pH 4.50, you'll need to use the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and the conjugate base.
The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
Where:
pH is the desired pH of the buffer (4.50 in this case)
pKa is the dissociation constant of the acid (4.76 in this case)
[A-] is the concentration of the conjugate base (sodium acetate in this case)
[HA] is the concentration of the acid (acetic acid in this case)
Let's assume x is the amount of sodium acetate (in moles) you need to add to the solution.
To begin, calculate the concentration of acetic acid:
[HA] = 0.100 M (since that's the initial concentration)
Now, we need to find the concentration of the conjugate base:
[A-] = x / 0.100 L (as we have added x moles of sodium acetate to a 100 mL solution)
Plug these values into the Henderson-Hasselbalch equation:
4.50 = 4.76 + log(x/0.100)
Rearrange the equation:
log(x/0.100) = 4.50 - 4.76
log(x/0.100) = -0.26
Take the antilog of both sides to remove the logarithm:
x/0.100 = 10^(-0.26)
Simplify:
x = 0.100 * 10^(-0.26)
Now, calculate x using a calculator:
x ≈ 0.0799 moles
So, you need to add approximately 0.0799 moles of sodium acetate to 100 mL of the 0.100 M acetic acid solution to obtain a buffer of pH 4.50.