You are given 255 mL of a 0.02 M base solution, and find out the pH of the solution is 10.4.
(a). Would this base be considered weak, strong, or negligible? Explain your answer.
(b). Determine whether the pH, pOH, % ionization, and equilibrium constant will increase, decrease, or stay the same when compared to the original solution.
-100 mL of water is added to the base solution.
pH:_____ % Ionization:_____ Kb:_____
-255 mL of 0.01 M of the same base is added to the solution.
pH:_____ % Ionization:_____ Kb:_____
Please explain your answers, thanks.
You are given 255 mL of a 0.02 M base solution, and find out the pH...
A 25.0-mL sample of 0.10 M weak base is titrated with 0.15 M strong acid. What is the pH of the solution after 9.00 mL of acid have been added to the weak base? Weak base Kb = 6.5 × 10–4
The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) this constant is given by Kb=[BH+][OH−][B] Strong bases will have a higher Kb value. Similarly, strong bases will have a higher percent ionization value. Percent ionization=[OH−] equilibrium[B] initial×100% Strong bases, for which Kb is very large, ionize completely (100%). For weak bases, the percent ionization changes with concentration. The more dilute the solution, the greater the percent ionization....
You have 15.00 mL of a 0.100 M aqueous solution of the weak base C5H5N (Kb = 1.50 x 10-9). This solution will be titrated with 0.100 M HCl. (a) How many mL of acid must be added to reach the equivalence point? (b) What is the pH of the solution before any acid is added? (c) What is the pH of the solution after 10.00 mL of acid has been added? (d) What is the pH of the solution...
You have 25.00 mL of a 0.100 M aqueous solution of the weak base CH3NH2 (Kb = 5.00 x 10-4). This solution will be titrated with 0.100 M HCl. (a) How many mL of acid must be added to reach the equivalence point? (b) What is the pH of the solution before any acid is added? (c) What is the pH of the solution after 5.00 mL of acid has been added? (d) What is the pH of the solution...
3. The pH of a 0.345 M weak base solution is 9.39. 50.0 mL of the weak base are titrated with 0.425 M HCI. a. Calculate the Kb of the weak base. b. Calculate the Ka of the weak base. c. d. Consult a Chemistry text and determine the identity of the weak base. Calculate the volume of HCl required to reach the equivalence point.
ANS:. . 4 3-What is the pH of a solution prepared by mixing 50 mL of 0.02 19 M ammonium chloride with 50 mL of 0.0292 M ammoni Report your answer to three decimal places ANS:. 4- Calculate the pH if you add 1 ml of 0.01 M HCI to the buffer from question #3. Repost you answer to three decimal places. ANS:.. 5-Calculate the pH if you added 1 mL of 0.01 M NaOH to the buffer from question...
During a titration of a weak base with a strong acid, you are slowing converting molecules of the weak base into molecules of its conjugate acid. For the hypothetical weak base, B we see the following: B (aq) + H30+ (aq) - BH+ (aq) + H20 (1) In the problem below you will be adding some strong acid, but not enough to reach the endpoint of the titration. 2.00 mL of hydrochloric acid added to the weak base 2.00M hydrochloric...
6. You have 20.00 mL of a 0.100 M aqueous solution of the weak base (CH3)3N (Kb = 7.40 x 10-5). This solution will be titrated with 0.100 M HCl. (a) How many mL of acid must be added to reach the equivalence point? (b) What is the pH of the solution before any acid is added? (c) What is the pH of the solution after 5.00 mL of acid has been added? (d) What is the pH of the...
Calculate the pH of the following solutions. a) 25.0 mL of 0.0035 M HClO (Ka of HClO= 2.9 x 10-8) b) 100 mL of 0.015 M CH3NH2(Kb of CH3NH2= 4.4 x 10-4) c) 500.0 μg/mL solution of Aniline, C6H5NH2. Aniline is a weak organic base with a pKb= 9.37. d) A 0.185 M solution of a weak base (B) has 2.04% ionization. Calculate the base dissociation constant, Kb, for the base.
Calculate the pH after titrating 50.0 mL of a 0.100 M weak base solution (Kb = 1.7 x10-9 ) with 50.0 mL of 0.200 M HBr