I am beyond confused about titration. Any help with work shown would be greatly
ID: 794716 • Letter: I
Question
I am beyond confused about titration. Any help with work shown would be greatly appreciated. Thank you!
Consider the titration of 33.0 mL of 0.117 M ammonia with 0.0700 M HCl.
I am beyond confused about titration. Any help with work shown would be greatly appreciated. Consider the titration of 33.0 mL of 0.117 M ammonia with 0.0700 M HCl. How many mL of HCl are required to reach the equivalence point? What is the pH at the equivalence point? What is the pH of the solution after the addition of 13.7 mL of acid? What is the pH of the solution after the addition of 86.2 mL of acid?Explanation / Answer
a) at equivalence point HCl moles = NH3 moles
0.07 x V = 0.117 x 33 , V = 55 ml
b) at equi valece point all NH3 is converted to NH4+ and hence back reaction occurs
NH4+ (aq) <-----> NH3 + H+
Ka = Kw/Kb(NH3) = 10^-14/10^-4.76 = 5.754 x10^-10
[NH4+] = ( 0.117x0.033 -x)/(0.055+0.033) = (0.00386-x)/(0.088) , [NH3]=[H+] =x/0.088 ,
Ka =[H+][NH3]/[NH4+] , 5.754 x10^-10 = ( x/0.088)^2/( 0.00386-x)/0.088
x = H+ moles = 4.42 x10^-7 , [H+] = ( 4.42 x10^-7)/( 0.088) = 5 x10^-6
pH = 5.3
c) HCl vol = 13.7 ml , then
NH3 + H+ <--> NH4+ , H+ moles =NH4+ moles = (0.0137x0.07) = 0.00096
NH3 = ( 0.117x0.033-0.00096) = 0.0029
pOH = pkb + log [NH4+]/[NH3] = 4.76+log(0.00096/0.0029) = 4.28 , pH = 14-4.28 = 9.72
d) when HCl vol = 86.2 ml we have excess H+ compared to base NH3
excess H+ moles = ( 0.07x0.0862 -0.117x0.033) = 0.002173
[H+]=(0.002173)/(0.0862+0.033) = 0.01823
pH = -log ( 0.01823) =1.74
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