You have the equation (CH3COOH = HAc; CH3COO^- = Ac^-).
HAc + NaOH ==> NaAc + H2O
You are starting with 0.1 M x 2 L = 0.2 mole HAc and 0.2 mole Ac^-.
We want 4.982
4.982 = 4.744 + log [(Ac^-)/(HAc)]
You must recognize that when NaOH is added, the HAc is decreased by x so the final moles = 0.2-x. At the same time, the amount of NaOH added (which is x) increases the Ac^- by x and it becomes 0.2+x. Substitute and solve for x.
4.982 = 4.744 + log [(0.2+x)/(0.2-x)]
Solve for x.
I don't know that carrying the pH to three places is justified so you need to check that out.
We have two liters of a buffer solution that
is 0.10 M in CH3COOH and 0.10 M in
NaCH3COO. How many moles of solid NaOH
are to be added to this solution to increase its
pH by 5.0 percent? Assume no volume change
due to the addition of solid NaOH.
I have the starting pH and what the final should be.
Initial would be 4.744 and final would be 4.98
1 answer