2. A buffer is created by combining 3.55 g of NH3 with 4.78 g of HCl and dilutin
ID: 897665 • Letter: 2
Question
2. A buffer is created by combining 3.55 g of NH3 with 4.78 g of HCl and diluting to a total volume of 750.0 mL.
a. Determine the pH of the buffer
Hint: First carry out a limiting reactant neutralization to get resulting moles of NH3 and NH4(+). Then use the "moles" form of the buffer equation to get the pH. No need to calculate molarities.
b. What is the pH after adding an additional 50.0 mL of 0.200 M NaOH to the buffer?
Hint: Again carry out a limiting reactant neutralization to get new moles of NH3 and NH4(+). Again used the "moles" form of the buffer equation to get the pH.
Explanation / Answer
m = 3.55 g of NH3
m = 4.78 g of HCl
VT = 750 ml
The reaction taking place
NH3 + HCl ---> NH4Cl
Find concentrations by first finding moles
MW of NH3 = 17 g/mol
MW of HCl = 36 g/mol
mol of NH3 = 3.55/17 = 0.2088 mol of NH3
mol of HCl = 4.78/36 = 0.1328 mol of HCl
Since stoichiometric ratios are 1:1
1 mol of HCl reacts with 1 mol of NH3
limiting reactant is HCl
0.1328 mol of HCl react with 0.1328 mol of NH3 to form 0.1328 mol of NH4Cl
Final concentration of NH3 = 0.2088 - 0.1328= 0.076 mol of NH3
Final concnetratin of HCl = 0.1328-0.1328 = 0 mol of NH3
Concnetration of NH4Cl and NH3:
M = mol/V
M = 0.076/0.75 = 0.1013 mol of NH3 per liter
M = 0.1328/0.75 = 0.1771 mol of NH4Cl per liter
This is what is in solutoin
NH3 + H2O ---> NH4+ and OH-
NH4Cl ---> NH4+ and Cl-
NH4+ is in both reactions, there is a common ion, this is a buffer
this is described with the Henderson Hasslelbach equation
pOH = pKb + log(HB+/HB)
pKb of NH3 = 4.75
pOH = 4.75 + log(0.1771/0.1013) = 4.99
pH = 14-pOH = 14-4.99 = 9.01
pH = 9.01
b)
After adding 0.2 M of NaOH and V = 50 ml
* Volume increases from 750 ml to 800 ml
* Concnetration of OH- ion increases!
Find the new concentration of OH- ions
[OH-] = mol/V
mol of OH- added = M*V = 50ml*.2 = 10 mmol of OH-
mol of OH- already in solution = 0.1771 mol
Total mol of OH- = 0.1771 mol + 10mmol = 0.1771 + 10/1000 = 0.1871 mol of OH-
[OH-] = mol/V = 0.1871/(0.800) = 0.2339 M
Recalculate NH3 concnetration
M1*V1 = M2*V2
0.1013*750 = M2*800
M2 = 750/800*0.1013 = 0.0945
Recalculate in pOH equation
pOH = pKb + log(HB+/HB)
pOH = 3.75 + log(0.2339/0.0945) = 4.14
pH = 14-pOH = 14-4.14 = 9.86
pH = 9.86
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