a. 15.00 ml of .200 M CH3COOH Ka = 1.8 x 10^-5.
b. 36.0 ml of .100 M malonic acid, (H2M) (Ka1 = 1.42 x 10^-3, ka2 = 2.01 x 10^-6
Please see the solution attached.
a. Lets check out what the acid and what the base in part a is. We have a base NaOH (compounds with OH- groups directly known to be bases) and we have a weak acid which is CH3COOH. The reason we know it's weak is because it has a Ka. We know that acids usually dissociate in water to give H+ ions. In case of weak acids the dissociation of the mother acid is partial and dissolving 1 mole of the acid would not necessarily give us 1 mole of H+ in solution.
Now in order to calculate the pH of the solution let's write the equation of the reaction down:
NaOH + CH3COOH ---> CH3COONa + H2O
When a strong base reacts with a weak acid the reaction always goes to completion to rid us of all the base (if the base is the limiting reagent). In this case, let's calculate the number of moles of base added. Molarity * volume =number of moles and so
Moles of base= 0.15*35*1e-3 (1e-3 =10 to the power of -3)=0.00525 moles of base.
Now we have to ...
This solution explains in 680 words how to determine pH in the reaction between an acid and a base by describing the relationship between an acid and a base. Calculations of molarity, concentration and pH are completed and attached in a .doc file.