An acid with a pKa of 4.73 has a concentration of 0.100 M. What is the...

An acid with a pKa of 4.73 has a concentration of 0.100 M. What is the pH of the acid? Give to two decimal places.

The pH of a 0.100 M solution of a weak acid is 3.15. What is the pKa of the acid? Give to two decimal places.

Expert Solution

PH = 1/2Pka -1/2logc

= 1/2*4.73-1/2log0.1

= 2.365-1/2*-1

= 2.365+1/2

= 2.365 +0.5

= 2.865 >>>>>>>>>answer

2.

PH = 1/2 Pka -1/2logC

3.15 = 1/2Pka -1/2log0.1

3.15 = 1/2Pka+ 1/2

1/2Pka = 3.15-0.5

1/2Pka = 2.65

PKa = 5.3 >>>>>>>.answer

queen_honey_blossom answered 1 year ago

Your acid has a pka of 7.54 and a concentration of 0.084M. You are given...

Your acid has a pka of 7.54 and a concentration of 0.084M. You are given 38.6mL of this acid, what is the pH of this initial solution? Calculate the pH of the solution after adding 8.7, 17.1, 30.5, and 33.3 mL of 0.100M NaOH. At the equivalence point, what volume of base is added and what is the pH?

50.0 ml of an acetic acid (CH3COOH) of unknown concentration is titrated with 0.100 M NaOH....

50.0 ml of an acetic acid (CH3COOH) of unknown concentration is titrated with 0.100 M NaOH. After 10.0 mL of the base solution has been added, the pH in the titration flask is 5.30. What was the concentration of the original acetic acid solution? (Ka(CH3COOH) = 1.8 x 10-5)

Your acid has a pka of 7.54 and a concentration of 0.084M. You are given 38.6mL...

Your acid has a pka of 7.54 and a concentration of 0.084M. You are given 38.6mL of this acid, what is the pH of this initial solution? Calculate the pH of the solution after adding 8.7, 17.1, 30.5, and 33.3 mL of 0.100M NaOH. At the equivalence point, what volume of base is added and what is the pH?

Your acid has a pka of 6.45 and a concentration of 0.079M. You are given 40.8mL...

Your acid has a pka of 6.45 and a concentration of 0.079M. You are given 40.8mL of this acid, what is the pH of this initial solution? Calculate the pH of the solution after adding 6.8, 15.3, 27.9, and 33.7 mL of 0.100M NaOH. At the equivalence point, what volume of base is added and what is the pH?

4-Nitrobenzoic acid has a pKa of 3.442. What is the pH of a 0.25 M solution...

4-Nitrobenzoic acid has a pKa of 3.442. What is the pH of a 0.25 M solution of 4-nitrobenzoic acid in pure water?

Given the following substances and their initial concentration: a) 0.100 M HNO3 f) 0.100 M HF...

Given the following substances and their initial concentration: a) 0.100 M HNO3 f) 0.100 M HF k) 0.100 M HC6H5O b) 55.5 M H2O g) 0.100 M HNO2 l) 0.100 M Ba(OH)2 c) 0.100 M NaOH h) 0.100 M CH3NH2 m) 0.00491 M HF d) 0.100 M C2H5NH2 i) 0.100 M C5H5N n) 0.100 M HOCl e) 0.100 M KNO2 j) 0.100 M CH3NH3Br o) 0.100 M CaCl2 Calculate the [H3O+] and the pH of each of the solutions.

part 1 - A 0.100 M solution of an acid, HA, has a pH=2.00. What is...

part 1 - A 0.100 M solution of an acid, HA, has a pH=2.00. What is the value of Ka of this acid? part 2 - We treat Kw as a fixed value, even though it, like all equilibrium constants, varies with temperature. The pH of pure water at 50 oC is 6.630. What is Kw at this temperature?

25.0 mL of a 0.100 M NH3 is titrated with a strong acid. 0.100 M HCl....

25.0 mL of a 0.100 M NH3 is titrated with a strong acid. 0.100 M HCl. Calculate the pH of the NH3 solution at the following points during the titration: (Kb= 1.8 x 10^-5) A. Prior to the addition of any HCl. B: After the addition of 10.5 mL of a 0.100 M HCl. C: At the equivilance point. D: After the addition of 3 mL of 0.100 M HCl. Show your work.

Consider the titration of 0.100 M HCl, a strong acid, with 0.100 M NH3, a weak...

Consider the titration of 0.100 M HCl, a strong acid, with 0.100 M NH3, a weak base: HCL(aq)+NH3(aq)---NH4+(aq)+H2O(l) Calculate the pH at the following points in the titration when 25.0 mL of 0.0100 M NH3 is titrated with 0.100 M HCL. A. 0.0 mL of HCl added B. 10.0 mL of HCl added C. At equivalence point D. 35.0 mL of HCL added

HCN has a pKa = 6.2 x 10-10. If a 50.0 mL of 0.100 M HCN...

HCN has a pKa = 6.2 x 10-10. If a 50.0 mL of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH a) after 8.00 mL base, b) at the halfway point of the titration, c) at the equivalence point. Answers are below, I need to know how to get to them step by step/conceptually Confirm: a) 8.49 , b) 9.21 c) 10.96

Question