9.50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 104) is titrated with a 0.10
ID: 1007085 • Letter: 9
Question
9.50.00 mL of 0.10 M HNO2 (nitrous acid, Ka = 4.5 × 104) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH solution is added, the pH in the titration flask will be: B. 3.35. C. 2.41. D. 1.48 E. 7.00 10. A 30 0-mL sample of o20 M HCOOH was titrated with 0.20 M NaOH. The following data were le of 0.20 M HCOOH was titrated with 0.20 M NaOH. The following data were collected during the titration. What is the Ka for HCOOH? mL of NaOH 5 added PH 3.07 3.47 4.07 4.77 A. 1.1 x 10-7 B. 1.7 x 104 C. 1.2 x 10 D. 4.9 × 10-11 E. None of these choices is correct. 11. Which of the following indicators would be the best to use when 0.050 M benzoic acid (G-6.6 × 10-5) is titrated with 0.05 M NaOH? A. bromphenol blue, pH range: 3.0-4.5 B. bromcresol green, pH range: 3.8-5.4 C. alizarin, pH range: 5.7-7.2 D. phenol red, pH range: 6.9-8.5Explanation / Answer
9. pka = -log(4.5*10^-4) = 3.35
No of mol of HNO2 = 50/1000*0.1 = 0.005 mol
No of mol of KOH = 0.0025 mol
pH = 3.35+log(0.0025/0.0025)
= 3.35
answer: B
10. pH = pka + log(salt/acid)
No of mol of HCOOH = 30/1000*0.2 = 0.006 MOL
nO OF mol of NaOH added = 5/1000*0.2 = 0.001 mol
3.07 = x+log(0.001/0.005)
pka = 3.77
ka = 10^-3.77 = 1.7*10^-4
answer: B
11. pka = -log(6.6*10^-5) = 4.18
pH = 7+1/2(4.18+log0.025) = 8.29
suitable indicator = phenolred
answer: D
16. answer: C because gas converts in to solution. disorderness decreases.
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