Asked by guess who
                A chemist wishes to prepare 250mL of a buffer that is pH = 4.50. Beginning with 100mL of 0.12 mol L^(-1) acetic acid and a supply of 0.10 mol L^(-1) NaOH, explain how this could be done. How much 0.20 mol L^(-1) NaOH must be added to this buffer to raise the pH to 5.1? If the same amount of 0.20 mol L^(-1) NaOH were added to 250 mL of deionized water, what would the new pH be? 
            
            
        Answers
                    Answered by
            DrBob222
            
    I couldn't guess.
How many millimoles acetic acid (HAc) do you have? That's 100 mL x 0.12M = 12.
How much NaOH must be added? That's 0.1 M * x mL = 0.1x
.........HAc + OH^- ==> Ac^- + H2O
initial..12.....0........0.......0
add............0.1x...............
change.-0.1x..-0.1x......0.1x....0.1x
equil..12-0.1x..0.......0.1x.....0.1x
pH = pKa + log(base)/(acid)
4.50 = pKa + log(0.1x)/(12-0.1x)
Solve for x = mL 0.1M NaOH to be added to form buffer of pH = 4.50.
I will leave the last two parts for you; the last one is strictly a strong base problem. The other one follows this same kind of procedure using the Henderson-Hasselbalch equation.
Post your work if you get stuck.
    
How many millimoles acetic acid (HAc) do you have? That's 100 mL x 0.12M = 12.
How much NaOH must be added? That's 0.1 M * x mL = 0.1x
.........HAc + OH^- ==> Ac^- + H2O
initial..12.....0........0.......0
add............0.1x...............
change.-0.1x..-0.1x......0.1x....0.1x
equil..12-0.1x..0.......0.1x.....0.1x
pH = pKa + log(base)/(acid)
4.50 = pKa + log(0.1x)/(12-0.1x)
Solve for x = mL 0.1M NaOH to be added to form buffer of pH = 4.50.
I will leave the last two parts for you; the last one is strictly a strong base problem. The other one follows this same kind of procedure using the Henderson-Hasselbalch equation.
Post your work if you get stuck.
                    Answered by
            guess who
            
    uno for the pKa would it simply be 4.50 as well?
    
                    Answered by
            DrBob222
            
    No. It's the pKa for acetic acid. Ka = 1.8E-5. pKa = -log Ka.
    
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