Answer to Question #138332 in General Chemistry for Jeanina Muñoz

Question

A buffer was prepared by dissolving 15 grams, if ammonium chloride in 100 ml solution prepared by diluting 15 ml of concentrated ammonia [28 % by mass with a density of 1.00
g/ml] to make 100 mL solution of dilute ammonia.
a. What is the ph if the buffer solution?
b. What will be the change in pH if 10 ml of 0.05M HCl is added to 25 ml of this buffer solution?
c. What will be the change in pH if 10 ml of 0.05 NaOH is added to 25 ml of this buffer?

Answer 1

Molar mass of NH4Cl = 53.5 g/mol

In 100 ml solution 15 g NH4Cl

              = (15/53.5) mole

 Mass of 15 ml solution =(15ml×density of solution)

                      =( 15×1 ) g

 Mass of NH3 = 28% of 15 g 

              =(28×15/100) g

              = 4.2 g

 Molar mass of NH3 = 17 g/mole

 4.2g NH3 =(4.2/17) mole

 100 ml NH3 solution containing (4.2/17) mole NH3

On mixing the total volume = 200ml

200 ml buffer contain =[(100×4.2)/(200×17)] mol NH3

                    =(2.1/17) mole

200ml buffer contain = (100×15)/(200×53.5) mol NH4Cl

                    = 15/(53.5×2) mol. NH4Cl

a)

pOH of this buffer, pOH = pKa + log (mole of salt/mole of base in buffer solution)

              Or, pOH = pKa + log (mol of NH4Cl/ mole of NH3)

              Or, pOH = 4.74+log [{15/(53.5×2)}/(2.1/17)]

              pOH = 4.68

   pH = pKw - pOH

       = 14 - 4.68

       = 9.315

 Hence, pH of the buffer solution is = 9.315

b)

25 ml buffer contain = [(15×25)/(200×2×53.5)] mole NH4Cl

                  =(15/856) mole NH4Cl

                   

 25 ml buffer contain = [(2.1×25)/(17×200)]

 Mole.

                   = (2.1/136) mole. NH3

                   

  

NH3 cinsumes HCl to produce NH4Cl

100 ml 0.05 M HCl = (0.05×10/1000) mole HCl

                 = 5×10^-5 mole HCl

After consumption NH3, remain = [(2.1/136)-(5×10^-4)] mol

                 = 0.0148 mole

NH4+ contain =(15/856)+(5×10^-4)

             = 0.018 mole

pH = pKw- pOH

   =14 - 4.74 - log(NH4+ contain/NH3 contain)

   = 9.26 - log(0.018/0.0148)

   = 9.175

   

Hence, pH of the buffer solution is =9.175

C)

NaOH consume NH4+ , producing (NH3+H2O)

0.05 M 10 ml NaOH contain = [(0.05×10)/1000] mole NaOH

                        = 5×10^-4 mol

                        

NH4+ remain = [(15/856)-(5×10^-4)] mole

            = 0.017 mole

Total amount of NH3 = [(2.1/136)+(5×10^-4)] mole

                   = 0.0159 mole

                   

pH = pKw- pOH

   =14 - 4.74 - log(NH4+ contant /NH3 contant)

     = 9.26 - log(0.017/0.0159)

     =9.23

      

Hence, pH of the buffer solution is =9.23