Question

Calculate the pH of the buffers prepared in the following ways:

A. By mixing 50.00 mL of 0.200 M ammonium nitrate, NH4NO3, with 25.00 mL 0.200 M NaOH

B. By neutralizing half of the acid in 100.00 mL of 0.200 M ammonium nitrate solution with 1.0 M NaOH

C. By mixing 60.00 mL of 0.400 M sodium formate (NaCHOO) with 40.00 mL 0.360 M HCl

D. By mixing 80.00 mL of 1.00 M NH3 with 40.00 mL of 1.00 M HCl

E. By mixing equal volumes of 1.00 M NH3 and 0.50 M HCl

Answer 1

NH4NO3 + NAOH ---> NH3NO3Na + H2O

subtract lowest mmol

10mmol     5mmol    

5mmol left 0mmol left 5mmol for NH3NO3NA

PH = PKA + log(base/acid) = 4.1+ log(5/5) = 4.1

b) since neutralizes half the acid, it is 1/2 volume needed to completely neutralize it(since 1:1 ratio)

so m1v1 =m2v2 ---> (100ml)(.200M) = 1.0M*v2 ---> v2 = 20ml so half is needed so v2 = 10ml

NH4NO3 mmol is 20mmol

NH3NO3Na will again be 10mmol so half

same as before PH = Pka -log(base/acid) = 4.1

c) 24mmol of NaCHOO and 14.4mmol of HCL

NaCHOO + HCL ----> CHOOH + NaCl

CHOOH has 14.4mmol

9.6mmol of NaCHOO left and

PH = Pka - log(base/acid) where ka = 1.8*10^-4 and pka = -log(ka)

PH = 3.74 -log(9.6mmol/14.4mmol) = 3.92

d) Ka of NH3 = 5.6*10^-10, pka = 9.25

NH3 + HCL ---> NH4+ + Cl-

40mmol of nh3 left

NH4+ has 40mmol as well

PH = Pka - log(base/acid) = 9.25 - log(40mmol/40mmol) = 9.25

e) For this one, I believe you need to do an ice chart to find the PH of the solution.

Pick random volume 1ml

NH3 + HCL ---> NH4+ +Cl-

1mmol NH3 .5mmol HCL so NH4+ has .5mmol

NH4 +H20 --->NH3 + H30+

.50   -----          0        0

doing it you get

x^2/.50 = 5.6*10^-10

x = .000016733

PH = -log(x) = 4.77