In the titration of 0.04 M weak base B of 50mL with 0.100 M HCl, what is the solution pH after 3 mL HCl is added? pka of the conjugate acid of base (B) is 7.9
In the titration of 0.04 M weak base B of 50mL with 0.100 M HCl, what...
19. The conjugate base salt to a weak acid (NaA) is titrated with 0.100 M HCl to its equivalence point. A 25.0 mL solution of a 0.200 M solution of the salt was titrated. The pK, for the unknown conjugate acid is 4.31. (a) Will the equivalence point be acidic or basic for this titration? i.e. pH less than 7.0 or greater than 7.0? (b) What is the volume in mL needed of HCl to reach the equivalence point? (c)...
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...
Consider the titration of the titration of 50.0 mL of 0.100 M acetic acid (HC2H2O2) with 0.100 M. The pka = 4.76. d. Determine the pH after 50.0 mL of titrant (NaOH) have been added. This is the equivalence point. All of the acid has been converted to its conjugate base, pH is determined by the equilibrium for the conjugate base
can i get help with this question please 3. Consider the weak base-strong acid titration of 25.00 mL of 0.100 M NH, with 0.100 M HCl (weak base, strong acid). The Ks for this weak base is 1.8 x 10 a. Calculate the pH of the solution after the addition of 10.0 mL HCL. b. What is the pH half-way to the equivalence point? c. Calculate the pH at the equivalence point. d. Calculate the pH after the addition of...
Consider the curve shown here for the titration of a weak base with a strong acid and answer each question. a. What is the pH and what is the volume of added acid at the equivalence point? b. At what volume of added acid is the pH calculated by working an equilibrium problem based on the initial concentration and Ks of the weak base? c. At what volume of added acid does pH = 14 - pka ? d. At what volume of added...
Consider the titration of 50.0 mL of 0.133 M NH3 (a weak base with Kb = 1.76 x 10-5 ) with 0.223 M HCl (a strong acid). Calculate the pH of the solution at each of the following points: 1. What is the pH of the solution before the titration is begun? 2. What is the pH of the solution after the addition of 15 mL of HCl? 3. What is the pH of the solution at the equivalence point?...
Question: In the figure below, titration curves for strong acid with strong base and weak acid with strong base are shown. Compare the shapes of these curves early in the titration for three different cases: titration of a strong acid, titration of a weak acid with a lower pKa, and titration of a weak acid with a higher pKa. Discuss with the class why the titration curve for weak acids increase more rapidly early in the titration than do stronger...
Titration of 25.00 mL of 0.100 M HCl with 0.100 M NaOH (strong acid, strong base): Answer the following questions: 4. Calculate the initial pH 5 Why is pH = 7 at the equivalence point? 6Why does the pH rise slowly at first, very rapidly near the equivalence point, and slowly after the equivalence point? 7. Why does it require 25.00 mL of NaOH to reach the equivalence point?
Consider the titration of 40.0 mL of 0.0600 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HCl. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 6.0 mL pH = (c) 12.0 mL pH = (d) 18.0 mL pH = (e) 24.0 mL pH = (f) 26.4 mL pH =