Question

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

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.

Ans :-

We know,

Molarity = Number of moles / Volume of solution in L

So,

Number of moles of NH₃ = Molarity x Volume of solution in L

= 0.200 M x 0.075 L

= 0.015 mol

Similarly,

Number of moles of HNO₃ = Molarity x Volume of solution in L

= 0.500 M x 0.017 L

= 0.0085 mol

Now, ICF table of the reaction of HNO₃ or H⁺ with NH₃ :-

..........................NH₃ ..................+............H⁺ ------------------> NH₄⁺

Initial (I)..............0.015 mol......................0.0085 mol..............0.0 mol

Change (C)........-0.0085 mol....................-0.0085 mol.............+0.0085 mol

Equilibrium (E).....0.0065 mol....................0.0 mol.....................0.0085 mol

Using Henderson-Hasselbalch equation in order to calculate the pH of basic buffer solution.

pOH = pK_b + log [Conjugate acid] / [Base]

pOH = - log K_b + log [NH₄⁺] / [NH₃]

pOH = - log 1.8 x 10^-5 + log 0.0085 / 0.0065

pOH = 4.74 + 0.1165

pOH = 4.86

Because,

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 4.86

pH = 9.14

Hence, pH = 9.14