for 5 stars please show ALL WORK and CLEARLY (a suggestion; solvtion could be pr

Question

for 5 stars please show ALL WORK and CLEARLY

(a suggestion; solvtion could be presented as a picture of solution on paper and upload if possible)

Vinegar is about 5% by weight acetic acid. The p/Ka for acetic acid is 4.74 (Ka = 10-4.74). What is the morality of acetic acid? Call this C Assuming that nothing else is in the vinegar besides water and the acetic acid, the electroneutrality equation (ENE) will be [H+] = [CH3 COO-] + [OH-] where [CH3 COO] is the acetate ion. We know that the solution will be acidic, so we can neglect [OH-] in the ENE which then becomes [H+] [CH3 COO-] If we set [CH3 COO-] = x then knowing that C= [CH3 COOH] + [CH3 COO-] make substitutions in the expression for Ka and Solve for [H+] and pH. alpha 0 and alpha 1 (the fractions of C for the species CH3 COOH and CH3 COO-. Using the Ka value for NH4+ (10-9 26), if we put a bit of ammonia in the vinegar and not change the pH, what would the fraction of the free-base form (NH3) be ? This is alpha 1. By adding base to the vinegar, what value would you have to raise the pH to if you wanted that fraction to be 0.5?

Explanation / Answer

C6h6 + 15/2o2 --> 3h2o + 6co2

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.