Determine the acid dissociation constant for a 0.010 M nitrous acid solution tha

Question

Determine the acid dissociation constant for a 0.010 M nitrous acid solution that has a PH of 2.70 Nitrous acid is a weak monoprotic acid and the equilibrium equation of interest is HNO_2(aq)+H_2O(l) H_3O+(aq) + NO_2-(aq). A) 5.0 times 10^-4 B) 4.0 times 10^-4 C) 8.0 times 10^-3 D) 2.0 times 10^-3 Arrange the following monoprotic adds in increasing order of: a) acid strength b) [H_3O+] in a 0.1 M solution of each acid; C) pH of the solutions A) acetic acid with K_a = 1.8 times 10^-5 B) benzoic acid with K_a = 6.5 times 10^-5 C) hypochlorous acid with K_a = 3.5 times 10^-8 D) hydrofluoric acid with K_a = 3.5 times 10^-4 What is the hydronium ion concentration of a 0.100 M acetic acid solution with a K_a = 18 times 10^-5? The equation for the dissociation of acetic acid is: CH_3CO_2H(aq) + H_2O(l) H_3O+(aq)+ CH_3CO_2-(aq). A) 4.2 times 10^-2 M B) 1.3 times 10^-3 M C) 1.3 times 10^-2 M D) 4.2 times 10^-3 M A tablet containing 500.0 mg of aspirin (acetylsalicylic acid or HC_9H_7O_4) was dissolved in enough water to make 100 mL of solution. Given that K_a = 3.0 times 10^-4 for aspirin, what is the pH of the solution? A) 5.08 B) 1.57 C) 3.52 D)2.54 What is the percent dissociation of a benzoic acid solution with pH = 2.59? The acid dissociation constant for this monoprotic acid is 6.5 times 10^-5. A) 1.5% B) 2.5% C)3.5% D)0.50% Calculate the pH of a 0.20 M H_2SO_3 solution that has the stepwise dissociation constants K_a1 = 1.5 times 10^-2 and K_a2 = 6.3 times 10^-8 A) 1.82 B) 2.52 C) 1.26 D) 1.32 Which of the following can be classified as a weak base? A) Neither CH_3NH_2 nor NH_2OH B) Both CH_3NH_2 and NH_2OH C) CH_3NH_2 D) NH_2OH What is the pH of a 0.30 M pyridine solution that has a K_b = 1.9 times 10^-9 ? The equation for the dissociation of pyridine is C_5H_5N(aq) + H_2O(l) C_5H_5NH+(aq) + OH-(aq)? A) 4.62 B) 9.38 C) 10.38 D) 8.72

Explanation / Answer

25

from pH find the [H3O+]

[H3O+] = 10-pH = 10-2.7 = 0.00199

from the given equation

if 1 mole of HNO2 dissociated it will give 1 mol of H3O+ and 1 mol of NO2-

so concentration of H3O concentration of NO2- 0.00199

concentration of HNO2 after dissociation

= intial concentration of HNO2 - concentration of H3O+

= 0.01 - 0.00199

= 0.00801 M

Ka = [H3O+] [NO2-] / [HNO2]

Ka = [0.00199] [0.00199] / [0.00801]

Ka =5 x 10^-4

option A is correct

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