Question

An 80.0 mL sample of 0.200 M ammonia is titrated with 0.100M hydrochloric acid. Kb for...

An 80.0 mL sample of 0.200 M ammonia is titrated with 0.100M hydrochloric acid. Kb for ammonia is 1.8 x 10-5. Calculate the pH of the solution at each of the following points of the titration: a. before the addition of any HCl. ______________ b. halfway to the equivalence point.______________ c. at the equivalence point._________________ d. after the addition of 175 mL of 0.100M HCl.____________

Homework Answers

Answer #1

a) Intially, there is only ammonia in the titration flask. Consider the dissociation of ammonia.

We can calculate the concentration of OH- based on the kb value and then the pOH value.

Since we know the pOH, we can calculate the PH now.

b) At the mid point half of the amount of ammonia present is neutralized. We apply this knowledge into Henderson-Hasselbach equation,

c) The volume of HCl added to reach the equivalence point can be calculated as follows.

At the equivalence point, HCl neutralizes all the ammonia present. The tiration flask contains the conjugate acid of ammonia, which is NH4+. The acidic salt undergoes hydrolysis as below,

The hydrolysis constant for the above reaction can be defined as,

The concentration of NH4+ can be calculated using,

We can calculate the concentration of H+ and then the pH using the equation for the hydrolysis of NH4+ .

d) The excess concentration of HCl,

The pH of the solution,

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1)A 40.0 mL sample of 0.045 M ammonia is titrated with 25.00 mL of 0.080 M...
1)A 40.0 mL sample of 0.045 M ammonia is titrated with 25.00 mL of 0.080 M HCl. If pK_bb​ = 4.76 for ammonia, what is the solution pH? 2) The single equivalence point in a titration occurs at pH = 5.65. What kind of titration is this? A Strong acid titrated with a strong base B Weak acid titrated with a strong base C Weak base titrated with a strong acid
A 46.3 mL sample of a 0.380 M solution of NaCN is titrated by 0.350 M...
A 46.3 mL sample of a 0.380 M solution of NaCN is titrated by 0.350 M HCl. Kb for CN- is 2.0×10-5. Calculate the pH of the solution: (a) prior to the start of the titration pH = (b) after the addition of 25.1 mL of 0.350 M HCl pH = (c) at the equivalence point pH = (d) after the addition of 71.9 mL of 0.350 M HCl. pH =
A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M...
A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M HCl. Kb for CN- is 2.0×10-5. Calculate the pH of the solution: (a) prior to the start of the titration pH = (b) after the addition of 48.4 mL of 0.260 M HCl pH = (c) at the equivalence point pH = (d) after the addition of 76.2 mL of 0.260 M HCl. pH =
Titration of Ammonia with Hydrochloric Acid a)Find the pH of 35.00mL of 0.100M ammonia b) Find...
Titration of Ammonia with Hydrochloric Acid a)Find the pH of 35.00mL of 0.100M ammonia b) Find the pH after 15.00mL of 0.100M HCL has been added to 35.00mL of 0.100M NH3
Determine the pH during the titration of 34.1 mL of 0.391 M ammonia (NH3, Kb =...
Determine the pH during the titration of 34.1 mL of 0.391 M ammonia (NH3, Kb = 1.8×10-5) by 0.391 M HNO3 at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem. (a) Before the addition of any HNO3 ? (b) After the addition of 15.0 mL of HNO3 ? (c) At the titration midpoint ? (d) At the equivalence point ? (e) After adding 54.2 mL...
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH....
A 20.00-mL sample of formic acid (HCO2H) is titrated with a 0.100 M solution of NaOH. To reach the endpoint of the titration, 30.00 mL of NaOH solution is required. Ka = 1.8 x 10-4 What is the pH of the solution after the addition of 10.00 mL of NaOH solution? What is the pH at the midpoint of the titration? What is the pH at the equivalence point?
2.)a.)A 24.1 mL sample of 0.370 M methylamine, CH3NH2, is titrated with 0.352 M nitric acid....
2.)a.)A 24.1 mL sample of 0.370 M methylamine, CH3NH2, is titrated with 0.352 M nitric acid. At the equivalence point, the pH is B.)A 27.8 mL sample of 0.243 M dimethylamine, (CH3)2NH, is titrated with 0.226 M nitric acid. The pH before the addition of any nitric acid is C.)A 23.2 mL sample of 0.389 M diethylamine, (C2H5)2NH, is titrated with 0.387 M hydrochloric acid. After adding 33.3 mL of hydrochloric acid, the pH is
1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with...
1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated with 0.102 M NaOH solution, what is the pH of the titration mixture after 13.7 mL of base solution is added? 2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with 0.108 M NaOH solution, what is the pH of the acid solution before any base solution is added? 3)If 27.9 mL of 0.107 M acid with a pKa of...
Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many...
Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the pH at the following points: a)After the addition of 10.0 mL of base b)Halfway to the equivalence point c)At the equivalence point d)After the addition of 80.0 mL of base
A 50.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is titrated with 0.150 M HNO3....
A 50.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is titrated with 0.150 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. A. 0.0 mL B. 25.0 mL C. 50.0 mL