Given: A 70.0 mL solution of 0.191 M ammonia is titrated with 0.191 M HCl. The...

Given: A 70.0 mL solution of 0.191 M ammonia is titrated with 0.191 M HCl. The pKa for ammonium is 9.3.

Question:

The titration is basically figuring out what is in the breaker and selecting the correct formula.

1) 0.0mL HCl Added

What is in the beaker: __________

What is the formula ? __________

Calculate the numbers:

Find the pH. :

2) 25 mL of HCl Added

What is in the beaker? _____________

What is the formula: ____________

The numbers come from stoichiometry of the titration reaction.

Put the numbers in:

pH = ?

Solutions

Expert Solution

queen_honey_blossom answered 4 months ago

Next > < Previous

Related Solutions

A volume of 70.0 mL of a 0.820 M HNO3 solution is titrated with 0.900 M...

A volume of 70.0 mL of a 0.820 M HNO3 solution is titrated with 0.900 M KOH. Calculate the volume of KOH required to reach the equivalence point.

A 50.0 mL solution of 0.200 M ammonia (pKa of ammonia is 9.24) is titrated by...

A 50.0 mL solution of 0.200 M ammonia (pKa of ammonia is 9.24) is titrated by 0.100 M HCl. Calculate the pH of the solution when, A.) the volume of HCl is 20.00 mL B.) the volume of HCl is 100.0 mL

A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution....

A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the addition of 20.0 mL of the HCl solution. (potentially useful info: Ka of NH4+ = 5.6 x 10−10)

1) Suppose 500 mL of an 0.200 molar HCl acid is titrated with 70.0 mL of...

1) Suppose 500 mL of an 0.200 molar HCl acid is titrated with 70.0 mL of a 1.00 M NaOH solution. What is the final pH of the solution? The error interval is 0.1 pH units. 2) Suppose 500 mL of 0.100 M benzoic acid, which has a pKa of 4.201, solution is titrated with 75.0 mL of a 1.00 M NaOH. What is the final pH? The error interval is +/- 0.2 pH units.

A 10.0−mL solution of 0.570 M NH3 is titrated with a 0.190 M HCl solution. Calculate...

A 10.0−mL solution of 0.570 M NH3 is titrated with a 0.190 M HCl solution. Calculate the pH after the following additions of the HCl solution a)0mL b)10ml c)30mL d)40mL

A 10.0-ml solution of 0.780 M NH3 is titrated with a 0.260 M HCL solution. calculate...

A 10.0-ml solution of 0.780 M NH3 is titrated with a 0.260 M HCL solution. calculate the pH after the following additions of the hcl solution: A. 0.00 B. 10.0 C. 30.0 D. 40.0

A 50.0 mL solution of 0.167 M KOH is titrated with 0.334 M HCl . Calculate...

A 50.0 mL solution of 0.167 M KOH is titrated with 0.334 M HCl . Calculate the pH of the solution after the addition of each of the given amounts of HCl . 0.00 mL pH = 7.00 mL pH = 12.5 mL pH = 19.0 mL pH = 24.0 mL pH = 25.0 mL pH = 26.0 mL pH = 31.0 mL pH =

A 25.00 mL solution of 0.100 M CH3NH2 is titrated with 0.1331 M HCl. What is...

A 25.00 mL solution of 0.100 M CH3NH2 is titrated with 0.1331 M HCl. What is the pH at the equivalence point of the titration? Species (K values) CH3NH2 (Kb = 3.74x10-4) CH3NH3+ (Ka = 2.70x10-11)

A 50.0 mL solution of 0.199 M KOH is titrated with 0.398 M HCl. Calculate the...

A 50.0 mL solution of 0.199 M KOH is titrated with 0.398 M HCl. Calculate the pH of the solution after the addition of the following amounts of HCl. a) 0.00 mL HCl b) 7.00 mL HCl c) 12.5 mL HCl d) 20.0 mL HCl

A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...

A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate 1) the concentration of the acid in the original solution, 2) the pH of the original HCl solution and the original NaOH solution, 3) the pH after 10.00 mL of NaOH have been added, 4) the pH at the equivalence point, and 5) the pH after 25.00 mL of NaOH have been...

Subjects