C. Detcrmination of pk,of Unknown Acid D. Concentration of Unknown Acid E NuCM M
ID: 554147 • Letter: C
Question
C. Detcrmination of pk,of Unknown Acid D. Concentration of Unknown Acid E NuCM Molarity of eknows as Sandsed desiation QUESTIONS 1. What are the largest sources of ermor in this esperinent? 2. What is the plt of the solution obeaised by mising 30.00 el, of 0.250 M HCI and 30.00 ml, of 0.125 Naoii? We assame additive volamses what is the pHE of a solution that i40 75 Arinsodium acetise and o so Ar in aceic acid?k,fr Velume at equivalence point Velune at one-hal equivalence poin pk, pk, Standard deviation of K,Explanation / Answer
For pKa measurement and concentration of acid we need molarity of original base solution, volume of acid solution taken.
Answers
1. In the given titration experiment, the largest source of error comes from incorrect measurement of molarity of base NaOH. This would give us incorrect moles of base used and hence the error in value of acid present in original solution. Also another major error comes from incorrect reading of level of buret for the volume of base added to the solution. The end point in these experiments can also be misread, say phenolphthalein used as indicator gives pink coloration in excess base. If the student adds excess base to get dark pink we would get incorrect volume for base added and hence the final value for molarity of acid present in solution would also be incorrect.
2. pH of solution
Titration of strong acid HCl and strong base NaOH
moles HCl present = 0.250 M x 30 ml = 7.5 mmol
moles NaOH added = 0.125 M x 30 ml = 3.75 mmol
excess [HCl] remained = (7.5 - 3.75) mmol/60 ml = 0.0625 M
pH = -log[H+] = -log(0.0625) = 1.204
3. This is a buffer solution
weak acid (Acetic acid)/conjugate base (sodium acetate)
using Hendersen-hasselbalck equation,
pH = pKa + log(base/acid)
= 4.74 + log(0.75/0.5)
= 4.92
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.