Take 50.00 mL (using your 25.00 mL pipet) of your vinegar solution, 25.00 mL of
ID: 496332 • Letter: T
Question
Take 50.00 mL (using your 25.00 mL pipet) of your vinegar solution, 25.00 mL of the NaOH solution (using your 25.00 mL pipet) and 25.00 mL of deionized water (using your 25.00 mL pipet) |and combine them in a labeled Erlenmeyer Flask. This solution is called AB1. (This gives 100 mL; keep it for Step 2) Obtain 10.00 mL (pipet) of solution AB1 and dilute it to 100.0 mL (volumetric flask) with deionized water. Ensure that it is thoroughly mixed. This is solution AB2. M NaOH: 0.10986 M plusminus 0.0012 M M Vinegar: 0.08591 M plusminus 0.003 M Table 1: Vinegar and NaOH Not sure how to calculate the (PH Calculated) if you could just show me how to do AB1 and AB2 I would appreciate it thank you.Explanation / Answer
pH calculated is the one you will get the the theoretical equations of analytical chemistry (i.e. titrations of weak acid + strong base)
so...
AB1:
Total V = 50 mL of vinegar + 25 mL of NaOH + 25 mL water = 100 mL
recalculate concentrations of each
M1*V1 = M2*V2 than:
Mnew = M*V / Vtotal
[Acetic acid] = M*V / (Vtotal) = 50 mL /100 mL *0.08591 M = 0.042955 mmol
[NAOH] = MV = 25*0.10986 / 100 = 0.027465 mmol
then... clearly, there is excesa acid so:
mmol of acetic acid lef tafter neutralization = 0.042955 - 0.027465 = 0.01549 mmol of acetic acid left
mmol of conjugate formed = 0 + 0.027465 = 0.027465 mmol of acetate left
so... this is clearly a buffer
it is desribed by Henderson HAsselbach equation
pH = pKa + log(Acetate/Acetic acid)
pKa = 4.75 for acetate/acetic acid
so
pH = 4.75 + log(0.027465 / 0.01549 ) = 4.99872
pH = 4.99872
your measurement was 4.85, so this is pretty near...
b)
from solution AB1 --> 10 mL
dilution to 100 mL ...
note that the concentrations are less....
but ratio of acetate/acetic ion is kept, since they are not reacting
so...
assume ratio AB1 = ratio AB2
therefore:
pH = 4.75 + log(0.027465 / 0.01549 ) = 4.99872
pH = 4.99872
note that you will expect this via calculation, but in reality, activity must be considered, which decreases as volume increases, so pH changes...
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