4. 100.0 mL of 0.0500 M NH3 is titrated with 0.100 M HCl. (30 pts) a. Calculate

Question

4. 100.0 mL of 0.0500 M NH3 is titrated with 0.100 M HCl. (30 pts)
a. Calculate the volume of HCl necessary to neutralize the NH3.
b. Calculate the pH at the equivalence point.
c. Calculate the pH when 0.00 mL of titrant is added.
d. Calculate the pH when 10.00 mL of titrant is added.
e. Calculate the pH when 20.00 mL of titrant is added.
f. Calculate the pH when 30.00 mL of titrant is added.
g. Calculate the pH when 40.00 mL of titrant is added.
h. Calculate the pH when 60.00 mL of titrant is added.
i. Calculate the pH when 70.00 mL of titrant is added.
j. Calculate the pH when 80.00 mL of titrant is added.

Explanation / Answer

The reaction is NH3+HCl-------> NH4Cl

1 mole of NH3 requires 1 mole of HCl for neutralization.

So moles of NH3 in 0.05M and 100ml =0.05*100/1000 =0.005 moles

Moles of HCl =0.005, volume of HCl =0.005/0.1 =0.05L= 50ml

volume of mixture = 100+50= 150ml

moles of NH4Cl =0.005 and its concentration =0.005*1000/(100+50)=0.0333M

At equivalence point, NH4Cl is present and pH is governed by hydrolysis of NH4Cl

NH4+ + H2O--------> NH3 + H3O+

Ka= [H3O+] [NH3+]/ [NH4+] = Kw/Kb= 10-14/(1.8*10-5)= 5.55*10-10

x2/ (0.0333-x)= 5.55*10-10, where x= [H3O+] when solved x= 0.0000043, pH= 5.36

Kb=x2/(0.05-x)= 1.8*10-5,x=0.00094, pOH= 3.02, pH= 14-3.03= 10.97

for 10ml of HCl, Moles of NH3= 0.05*100/1000 =0.005, moles in .1M of 10ml HCl =0.1*10/1000=0.001 So NH3 is excess and moles of NH3= 0.005-0.001= 0.004

Volume after mixing = 100+10=110ml =0.11L

Concentration of NH3= 0.004/0.11 = 0.036 M

NH3+ H2O-------à>NH4+ OH-

Kb= [NH4+] [OH-]/ [NH3] = x2/(0.036-x)= 1.8*10-5, x =[OH-]

When solved x=0.000766, pOH= 3.116, pH= 14-3.116=10.884

Volume after mixing = 100+20=120ml =0.12L

Concentration =0.003/0.12=0.025M

X2/(0.025-x)= 1.8*10-5, x=0.000664, pOH= 3.17, pH= 14-3.17=10.83

Volume after mixing =130ml=0.13 L

Concentration of NH3= 0.002/0.13=0.0154

Therefore x2/(0.0154-x)= 1.8*10-5 x =0.000518, pOH= 3.28, pH= 14-3.28= 10.72

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