A titration of 60.0mL of 0.30M pyridine, C5H5N, required 90.0mL of 0.20 M HCl to reach the equivalence point. What is the pH at this equivalence point? Kb(C5H5N) = 2.0x10^-9
We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
A titration of 60.0mL of 0.30M pyridine, C5H5N, required 90.0mL of 0.20 M HCl to reach...
4. Suppose you titrate 50.00 mL of 1.20 M pyridine, C5H5N, with 0.634 M HCl. (a) How many moles of pyridine are present in the original pyridine solution? (b) What is the initial pH? (c) What volume (in mL) of HCl solution is required to reach the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH of the solution after the addition of 104 mL of HCl?
4. Suppose you titrate 50.00 mL of 1.20 M pyridine, C5H5N, with 0.634 M HCl. (a) How many moles of pyridine are present in the original pyridine solution? (b) What is the initial pH? (c) What volume (in mL) of HCl solution is required to reach the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH of the solution after the addition of 104 mL of HCl?
Calculate the pH of a 0.20 M solution of the weak base pyridine. (C5H5N; Kb = 1.7 x 10-9). What is the pH of a 0.20 M solution of NH4Cl? [Kb(NH3) = 1.8 ´ 10–5]
1))))Weak Base Titration: At the Equivalence Point a))A 11.30 mL solution of 2.47 M pyridine (C5H5N, a weak base) is titrated to the equivalence point with 22.4 mL of HNO3. What is the pH at the equivalence point? Kb for pyridine is 1.7 X 10-9. b))))Click on all of the dominant species that you would expect to find in solution at the equivalence point. Do not include H3O+ and OH-ions unless they are coming from another source other than the...
Thank you in advance. A titration of 50 ml of 0.20 M methyl amine, CH3NH2, required 50 ml of 0.20 M HCL to reach the equivalence point. what is the pH at this endpoint? Kb for CH3NH2 = 4 x 10^-4
A chemist titrates 60.0mL of a 0.3735M pyridine (C5H5N) solution with 0.6999M HNO3 solution at 25°C. Calculate the pH at equivalence. The pKb of pyridine is 8.77. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.
Consider the titration of 25.00 mL of 0.08364 M pyridine with 0.1067 M HCl (a) What volume of the titrant must be added to reach the equivalence point? (b) Find the pH when 4.63 mL of the titrant has been added.
Find the [OH−] of a 0.40 M pyridine (C5H5N) solution. (The value of Kb for pyridine (C5H5N) is 1.7×10−9.) Find the pH of a 0.40 M pyridine (C5H5N) solution.
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...
30.0 mL of 0.340M C5H5N (Pyridine Kb = 1.7 x 10-9) is titrated with 0.190M HBr (hydrobromic acid). Determine the pH at each of the following points. a. Before any HBr is added b. After 3.00mL of HBr is added c. at half titration d. at equivalence e. After 0.50 mL of HBr added after the equivalence point