Answer to Question #83448 in General Chemistry for natasha

Question

500.0 ml of a buffer solution is 0.619 M in NH3 and 0.305 M in NH4Cl. 500.0 ml of a 0.218 M solution in HCl are added. The final volume is 1.0000 L. The Kb value for the NH3 is 1.8 x 10-5. What is the final pH of this solution?

Answer 1

Answer on Question #83448, Chemistry/ General Chemistry

500.0 ml of a buffer solution is 0.619 M in NH3 and 0.305 M in NH4Cl. 500.0 ml of a 0.218 M solution in HCl are added. The final volume is 1.0000 L. The Kb value for the NH3 is 1.8 x 10-5. What is the final pH of this solution?

Solution

Hydrochloric acid will react with ammonia, NH₃, a weak base, to produce the ammonium cation, NH₄⁺, the ammonia's conjugate acid , and water:

HCl(aq) + NH₃(aq) -> NH₄⁺ (aq) + H₂O

c=n/V, n= c*V

n(NH₃) = 0.619*0.5= 0.3095 moles

n(NH₄⁺) = 0.305*0.5 0.1525 moles

n(HCl) = 0.218*0.5 = 0.109 moles

HCl and NH₃ react in 1:1 mole ratio , then the resulting solution will contain:

n(HCl) = 0 moles (completely consumed)

n(NH₃) = 0.3095 - 0.109 = 0.2005 moles

n(NH₄⁺) = 0.1525 + 0.109 = 0.2615 moles

The final volume of the solution is 1.0000 L. The new concentartions of the ammonia and ammonium cations will be:

[NH₃] = 0.2005 moles/ 1.0000 L = 0.2005 M

[NH₄⁺] = 0.2615 moles/ 1.0000 l = 0.2615 M

Find pH by using Henderson-Hasselbalch equation? which for a buffer that contains a weak base and its conjugate acid look like this:

pOH = pKb + log ([conjugate acid]/[weak base])

pOH = -log(1.8*10^-5) + log(0.2615/0.2005) = 4.860

pH = 14-pOH

pH = 14 -4.860 = 9.140

Answer: 9.140