Question

In: Chemistry

A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl. Calculate the...

A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl.

Calculate the pH at one-half equivalence point.

Calculate the pH at 40 mL of added acid.

Calculate the pH at 50 mL of added acid.

Expert Solution

1)

millimoles of pyridine = 25.0 x 0.125 = 3.125

concentration of HCl = 0.100 M

Kb = 1.7 x 10^-9

pKb = 8.77

a) at half - equivalence point :

pOH = pKb

pOH = 8.77

pH = 5.23

b) at 40 mL of added acid.

mmoles of HCl = 40 x 0.100 = 4

C5H5N + HCl ----------------> C5H5NH+

3.125 4 0

0 0.875 3.125

here strong acid remains.

[H+] = 0.875 / 40 + 25 = 0.01346 M

pH = -log (0.01346)

pH = 1.87

c) 50 mL of added acid.

pH = 1.60

queen_honey_blossom answered 1 year ago

A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl. Part A...

A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl. Part A Calculate the pH at 0 mL of added acid. Express your answer using two decimal places. Part B Calculate the pH at 10 mL of added acid. Express your answer using two decimal places. Part C Calculate the pH at 20 mL of added acid. Express your answer using two decimal places. Part D Calculate the pH at equivalence point. Express your answer using...

A 25.0 mL sample of 0.100 M acetic acid is titrated with a 0.125 M NaOH...

A 25.0 mL sample of 0.100 M acetic acid is titrated with a 0.125 M NaOH solution. Calculate the pH of the mixture after 10, 20, and 30 ml of NaOH have been added (Ka=1.76*10^-5)

Consider the titration of a 25.0 mL sample of 0.170 molL−1 CH3NH2 (Kb=4.4×10−4) with 0.150 molL−1...

Consider the titration of a 25.0 mL sample of 0.170 molL−1 CH3NH2 (Kb=4.4×10−4) with 0.150 molL−1 HBr. Determine each quantity: Part A the volume of added acid required to reach the equivalence point Part B the pH at 6.0 mL of added acid Express your answer using two decimal places. Part C the pH at the equivalence point Express your answer using two decimal places. Part D the pH after adding 6.0 mL of acid beyond the equivalence point Express...

A 0.2000M pyridine (Kb=1.710^-9) solution with initial volume 10.000ml is titrated with 2.50010^-1 M HCL. what...

A 0.2000M pyridine (Kb=1.7*10^-9) solution with initial volume 10.000ml is titrated with 2.500*10^-1 M HCL. what is the pH after the addition of : a.) 2.00ml of HCL solution b.) 6.00ml of HCL solution c.)8.00ml of HCL solution d.)10.00ml of HCL solution

A 25.0 mL sample of 0.100 M HC2H3O2 (Ka = 1.8 x 10^-5) is titrated with...

A 25.0 mL sample of 0.100 M HC2H3O2 (Ka = 1.8 x 10^-5) is titrated with a 0.100 M NaOH solution. What is the pH after the addition of 12.5 mL of NaOH?

A 111.4 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.245 M HNO3. Calculate...

A 111.4 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.245 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. 0.0mL 22.7mL 45.5mL 68.2mL

A 100.0 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.250 M HNO3. Calculate...

A 100.0 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.250 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. Part A: 0.0 mL Part B: 20.0 mL Part C: 40.0 mL Part D: 60.0 mL

A 115.2 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.265 M HNO3. Calculate...

A 115.2 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.265 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. 43.5ml- 65.2 ml-

Consider the titration of a 21.0 mL sample of 0.110 molL−1 CH3COOH (Ka=1.8×10−5) with 0.125 molL−1...

Consider the titration of a 21.0 mL sample of 0.110 molL−1 CH3COOH (Ka=1.8×10−5) with 0.125 molL−1 NaOH. Determine each quantity: a) the initial pH. Express your answer using two decimal places. b) the volume of added base required to reach the equivalence point c) the pH at 4.0 mL of added base. Express your answer using two decimal places. d) the pH at one-half of the equivalence point. Express your answer using two decimal places. e) the pH at the...

Consider the titration of a 22.0 mL sample of 0.100 molL−1 CH3COOH (Ka=1.8×10−5) with 0.120 molL−1...

Consider the titration of a 22.0 mL sample of 0.100 molL−1 CH3COOH (Ka=1.8×10−5) with 0.120 molL−1 NaOH. Determine each quantity: Part A the initial pH pH = 2.87 Part B the volume of added base required to reach the equivalence point V = 18.3 mL Part C the pH at 5.0 mL of added base Express your answer using two decimal places. pH = Part D the pH at one-half of the equivalence point Express your answer using two decimal...