You begin with a beaker containing 50.0 mL of a solution of an unknown weak base
ID: 514163 • Letter: Y
Question
You begin with a beaker containing 50.0 mL of a solution of an unknown weak base with a concentration of 0.35 M and a K_b value of 4.0 times 10^7. You titrate this weak base with 0.300 M HCL. a. What is the pH of the solution before any HCL is added? b. How many mL of HCL are required to reach the halfway point? c. What is the pH at the equivalence point? d. What is the pH of the solution after 10.0 mL of HCI have been added? e. What is the pH of the solution after 75.0 mL of HCL have been added?Explanation / Answer
4. a) POH for a weak base=1/2 Pkb-1/2logc Putting the values we get POH=1/2(-log4.0*10^-7)-1/2log 0.35
=1/2(7-log4)-1/2log(0.35)=3.2+1-1/2log3.5=4.3+0.272=4.572
Thus PH=14-4.572=9.428
b) At half neutralisation point half of the base is neutralised. In this condition [salt]=[base] Thus using Henderson Heselbach equation for base buffer we get POH=Pkb+log[salt]/[base]=Pkb=6.4
PH=14-6.4=7.6
c) At equivalence point PH is the PH of the salt(made with weak base and strong acid) hydrolysis.
PH=7-1/2PKb-1/2logc=7-3.2-1/2log(0.35)=4.02
d) when 10 ml HCl will be added then a buffer will be produced .So using Henderson, Heselbach equation we get
POH=Pkb+log[salt]/[base]=6.4+log{[10*0.3]/[50*0.35-10*0.3]}=6.4+log3/14.5=5.71
So PH=14-5.71=8.29
e) When 75 ml HCl is added then [H+]=(75*0.3-50*0.35)/125=5/125
PH=-log(0.04)=2-log4=2-0.3010*2=1.4
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