Question

Part A: How many grams of dry NH4Cl need to be added to 2.40 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.71? Kb for ammonia is 1.8×10^−5.

Part B:

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 17.0 mL of HNO3.

Part C:

Barium sulfate, BaSO4, is used in medical imaging of the gastrointestinal tract because it is opaque to X rays. A barium sulfate solution, sometimes called a cocktail, is ingested by the patient, whose stomach and intestines can then be visualized via X-ray imaging. If a patient ingests 380 mL of a saturated barium sulfate solution, how much toxic Ba2+ ion has the patient consumed?

The solubility product Ksp of BaSO4 is 1.10×10^-10

Express your answer to three significant figures and include the appropriate units. Enter in units of Mass.

Answer 1

Here is the Solution:

(1). We need to to find out the mass in gram of dry NH4CI need to be added to 2.40 L of a 0.500 M solution of ammonia (NH3), to prepare a buffer solution that has a pH of 8.71 o We have 2.40 L of a 0.500 M solution of ammonia (NH3) o pH of the solution is: 8.71 o Kn for ammonia is (NH2) is: 1.8x 10 5 owe know that, the molar mass of ammonium chloride (NH4CI) is: 53.491 g/mol. (i). First, we will find out the acid concentration of the 2.40 L solution by using "Henderson Hasselbach" equation. acid рон 3DрКb + lоg base pOH 14-pH pOH 14 -8.71 рон 3D5.29 pKb log(Kb) pKb log(1.8 x 10-5) pKb 4.74 Now acid рон 3DрКb + lоg base acid 5.29 4.74 log 0.500 M mol acid 1.75486 M » This is the acid concentrat ion. (li) Second, we will find the number of moles of the conjugate acid present in 2.40 L of solution. mol 1.75486 x2.40= 4.21 166 mol. Therefore, the mass of ammonium chloride need to be added: Mass moles x molar mass Mass 4.21166 pmol x 53.491 g/paol Mass 225.2859 g Mass 225.29 grams

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