A flask containing 36.8 mL of 0.138 M sodium acetate (NaC2H3O2) is titrated with
ID: 489277 • Letter: A
Question
A flask containing 36.8 mL of 0.138 M sodium acetate (NaC2H3O2) is titrated with 24.9 mL of 0.128 M HNO3. Calculate the pH of the solution. pKa for acetic acid = 4.76. (a) 4.53 (b) 9.47 (c) 4.98
In the titration of 30.0 mL of a 0.200 M solution of a hypothetical compound NaH2M, what is the pH of the solution after the addition of 60.0 mL of 0.100 M NaOH? For H3M, pKa1 = 3.00, pKa2 = 6.00, and pKa3 = 9.00. (a) 6.00 (b) 9.00 (c) 7.50
For the titration of 10.00 mL of a 0.100 M nicotine (pKb1 = 6.15 and pKb2 = 10.85) with 0.100 M HCl, calculate the initial pH. (a) 10.42 (b) 13.00 (c) 3.58
What two assumptions must be true for the pH of an intermediate of a diprotic acid to equal (pKa1 + pKa2)/2? Overall 70 57 (a) Ka2F >> Kw and Ka1F << F (b) Ka2F << Kw and Ka1F << F (c) Ka1F >> Kw and Ka2F << F
Explanation / Answer
1)(c) 4.98
pH= pKa + log [salt/acid]
= 4.757 + log [0.138*36.8/61.7 /0.128*24.9/61.7]
= 4.96
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.