Question

How many grams of dry NH4Cl need to be added to 1.50 L of a 0.500 M solution of ammonia, NH3,to prepare a buffer solution that has a pH of 8.79? Kb for ammonia is 1.8*10^-5.

Answer 1

From Henderson Hesselbach equation :

       pOH = pK_b + log [ conjugate base ] / [acid]                .............. (1)

Initial moles of  NH₃ present in 1.50 L of 0.500 M = 1.50 L * 0.500 mol / L = 0.75 mol

     NH₃ + H₂O -----> NH₄⁺ + OH^-

     So,      p0H = pK_b + log [ NH₄⁺] / [NH₃]

    14.00 - 8.79  = - log(1.8*10^-5) + log ( x / 0.75 )         { where x is moles of NH₄⁺}

                 5.21 = 4.75 + log ( x / 0.75 )

                 0.46 = log ( x / 0.75 )

             x / 0.75 = 10^0.46 = 2.88

            

Then, moles of NH₄⁺ = 2.16 mol

Then mass of NH4Cl will be , 2.16 mol * ( 53. 49 g / 1mol ) = 116 g

Answer 2

From Henderson Hesselbach equation :

       pOH = pK_b + log [ conjugate base ] / [acid]                .............. (1)

Initial moles of  NH₃ present in 1.50 L of 0.500 M = 1.50 L * 0.500 mol / L = 0.75 mol

     NH₃ + H₂O -----> NH₄⁺ + OH^-

     So,      p0H = pK_b + log [ NH₄⁺] / [NH₃]

    14.00 - 8.79  = - log(1.8*10^-5) + log ( x / 0.75 )         { where x is moles of NH₄⁺}

                 5.21 = 4.75 + log ( x / 0.75 )

                 0.46 = log ( x / 0.75 )

             x / 0.75 = 10^0.46 = 2.88

            

Then, moles of NH₄⁺ = 2.16 mol

Then mass of NH4Cl will be , 2.16 mol * ( 53. 49 g / 1mol ) = 116 g

Answer 3

From Henderson Hesselbach equation :

       pH = pK_b + log [ conjugate base ] / [acid]                .............. (1)

Initial moles of  NH₃ present in 1.50 L of 0.500 M = 1.50 L * 0.500 mol / L = 0.75 mol

     NH₃ + H₂O -----> NH₄⁺ + OH^-

So, pH = pK_b + log [ NH₄⁺] / [NH₃]

    8.79  = - log(1.8*10^-5) + log ( x / 0.75 )         { where x is moles of NH₄⁺}

    8.79 = 4.74 + log ( x / 0.75 )

          x = 8.4 * 10³ mol

Then mass of NH4Cl will be , 8.4 * 10³ mol * ( 53. 49 g / 1mol ) = 4.5*10⁵ g