If there is still not enough information just putting the steps down would help
ID: 496328 • Letter: I
Question
If there is still not enough information just putting the steps down would help
Combine 50.00 mL (using your 25.00 mL pipet) of your vinegar solution and an appropriate amount of your NaOH solution (you may use a graduated cylinder) such that the acetic acid is completely neutralized with no excess NaOH present. What is this volume? You need to calculate this now based on the molarity of the vinegar and the NaOH. Show this calculation in your report. This is solution X. Take 50.00 mL (using your 25.00 mL pipet) of your vinegar solution and dilute it with deionized water (using a graduated cylinder) in the amount of that calculated volume from above. This is solution Y. Not sure how to calculate the (PH Calculated) if you could just show me 1 example I would appreciate it thank you.|Explanation / Answer
Solution X : Gives NaOAc
Solution Y : Gives AcOH
Beaker A:
Volume of NaOH used for neutralization = 0.08591 M x 50 ml/0.10986 M = 39.1 ml
Initial concentration of NaOAc (solution X) = 0.08591 M x 50 ml/(50 + 39.1) ml = 0.04821 M
1 ml of this solution X was mixed with 24 ml of solution Y
Final concentration of NaOAc in beaker A = 0.04821 M x 1 ml/25 ml = 0.00193 M
24 ml of solution Y is mixed with 1 ml solution X (AcOH)
final concentration of Y in beaker A = 0.08591 M x 24 ml/25 ml = 0.003436 M
pH of solution calculated by Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
pKa of AcOH = 4.74
So,
pH = 4.74 + log(0.00193/0.003436)
= 4.49
[pl. note: for solution Y, the dilution volume is not mentioned above (experimental details) and thus the actual concentration of solution Y would be different than the one used in the above shown calculations]