3)_ Calculate the pH's before and after adding 0.5mmol HCl to 100ml 0.02 Lactic acid + 0.08M Sodium Lactate mixture?
Let's call lactic acid HL and sodium lactate NaL. NaLis the base and HL is the acid.The Henderson-Hasselbalch equation is
pH = pKa + log (NaL/HL) = pKa + log (b)/(a)
pH = pKa + log (0.08/0.02) = ? You will need to look up the pKa for lactic acid or calculate it from pKa = -log Ka. This is the pH of the buffer as is.
Now for the addition of HCl. Here I will switch to millimoles instead of concentrations in mols/L (or millimoles/mL)
millimoles HL = 100 mL x 0.02 M = 2
millimoles NaL = 100 mL x 0.08 M = 8
adding HCl is 0.5 mmol
........................L^- + HCl ==> HL + Cl^-
I........................8.....................2..................
add.............................0.5...........................
C....................-0.5.....-0.5........+0.5
E......................7.5.........0............2.5
You have two routes from here. Both give you the same answer. One is technically not right but it's a lot shorter. Depends on how picky your prof is.
Summary:
mmols HL = 2.5 and (HL) = mmols/mL = 2.5/100 = 0.025
mmols L^- = 7.5 and (NaL) = 7.5/100 = 0.075
Technically pH = pKa + log (0.025/0.075) = ? where I've used concentrations but
pH = pKa + log (2.5/7.5) = ? which is the same answer BECAUSE the volume is the same (100 mL) and the volume will always cancel. On this problem it makes little difference because calculating the concentration by mmols/mL isn't a problem since dividing by 100 is so easy but in other problems the volume may be something like 174 mL or 190 mL or whatever in which case using millimoles is always easier since that funny volume in the denominator always cancels.