16.72g of sodium acetate trihydride (MW=136 g/mol) is dissolved in 100mL water. To this, 4.00g glacial acetic acid (Ka=2.2x10-5, MW=60.0 g/mol) is added, and distilled water is added to to the 500mL mark. Find the final pH and the molar concentration of the buffer.
What's wrong with using the Henderson-Hasselbalch equation.
By the way, you mean sodium acetate trihydrate
16.72g/molar mass = moles
moles/L = M (the volume you want to use is 0.5L (and not the 100 ml).
4.00g/molar mass
moles /L = M (again use 0.5L for final volume an not some other number).
Then pH = pKa + log (base/acid)
To find the final pH and molar concentration of the buffer, we can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the molar concentration of the acid and its conjugate base.
First, let's calculate the number of moles of sodium acetate trihydrate (NaCH3CO2) and glacial acetic acid (CH3COOH) that are added to the solution.
Number of moles of NaCH3CO2 = mass / molecular weight
= 16.72g / 136 g/mol
= 0.123 mol
Number of moles of CH3COOH = mass / molecular weight
= 4.00g / 60.0 g/mol
= 0.067 mol
Next, since the sodium acetate trihydrate dissociates into sodium acetate (NaCH3CO2) and water (H2O), the concentration of sodium acetate (CNaCH3CO2) can be calculated as follows:
CNaCH3CO2 = moles / volume
= 0.123 mol / 0.5 L
= 0.246 M
The concentration of acetic acid (CCH3COOH) remains the same since glacial acetic acid does not dissociate significantly.
Now we can plug these values into the Henderson-Hasselbalch equation:
pH = pKa + log (CNaCH3CO2 / CCH3COOH)
where pKa is the negative logarithm of the acid dissociation constant (Ka).
pH = -log(2.2x10^(-5)) + log(0.246 / 0.067)
We can use a calculator to evaluate the logarithmic terms:
pH = 4.66 + log(3.67)
pH = 4.66 + 0.563
pH = 5.22
Therefore, the final pH of the buffer solution is 5.22.
To find the molar concentration of the buffer, we need to sum the moles of both the acid (CH3COOH) and its conjugate base (NaCH3CO2):
Total moles of buffer = moles of CH3COOH + moles of NaCH3CO2
= 0.067 mol + 0.123 mol
= 0.190 mol
Molar concentration of the buffer (CB) is given by:
CB = moles / volume
= 0.190 mol / 0.5 L
= 0.380 M
Therefore, the molar concentration of the buffer is 0.380 M.