What is the pH of a propanoic acid/sodium propionate buffer after the addition of 10.0 mL of 1.0 M NaOH?
helpful:
pH = pKa + log (base/acid)
Ka=1.3x10^-5
Answer:
Your solution has 0.75 mol of HOAc and 0.60 mol of AcO(-).
If you add 0.01 mol of NaOH (10 mL of 1.0 M solution), then your final concentrations will be 0.74 mol of HOAc and 0.61 mol of AcO(-). Plug those into your buffer equation.
moles of OH- added = 0.01L*1.0M = 0.01 moles
new conc. of HC3H5O = (0.75mol - 0.01mol) / (1L + 0.01L)
new conc. of HC3H5O = 0.733M
new conc. of NaC3H5O2 = (0.60mol + 0.01mol) / (1L + 0.01L)
new conc. of NaC3H5O2 = 0.604M
pH = -log(1.3x10^-5) + log(0.604/0.733)
pH = 4.8
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