the complete solution to the problem I began in class, Wednesday Feb. 13:
15.0 mL of HClO4 (at 1.0 M) and 200 mL of NaOH (at 0.25 M) are mixed then diluted to a final volume of 500 mL.
Is the final solution acidic or basic? What is the pH?
Step 1. Write the reaction and identify the acid and base as strong or weak.
However, the number of moles of acid is not the same as base.
In other words, this is really a limiting reagent problem: which will be in excess?
HClO4 + NaOH ---> [Na+] + [ClO4-] + H2O
Both acid and base are strong so reaction will go to completion.
HClO4 + NaOH ---> [Na+] + [ClO4-] + H2O
x moles + y moles
# moles HClO4 = (15.0 mL x 1.0 M)/ (1 L/1000 mL) = 0.015 moles HClO4
# moles NaOH = (200 mL x 0.25 M)/ (1 L/1000 mL) = 0.050 moles NaOH
So, NaOH is in excess. We can expect there to be excess hydroxide, OH-, and for the final solution to be basic, pH >7.
Step 3. Calculate how much acid reacts to form the products, [Na+] and [ClO4-], which are spectator ions.
Then calculate the amount of excess, unreacted NaOH.
HClO4 + NaOH ---> [Na+] + [ClO4-] + H2O
.015 mol + .015mol ---> 0.015 moles [Na+] and [ClO4-]
Unreacted NaOH = 0.050 moles - 0.015 moles = 0.035 moles NaOH left
Step 4. Calculate the concentrstion of hydroxide formed from the unreacted NaOH. Convert molarity, M, to pOH.
NaOH ---> [Na+] + [OH-]
[OH-] , M = #moles left / final solution volume = 0.035 moles/ 0.500L
[OH-] = 0.0700 M
pOH = -log( 0.0700) = 1.15