Consider the titration of 33.0 mL of 0.117 M ammonia with 0.0740 M HCl. Use the
ID: 956947 • Letter: C
Question
Consider the titration of 33.0 mL of 0.117 M ammonia with 0.0740 M HCl. Use the Acid-Base Table. Plase see Titration for assistance.
(a) How many mL of HCl are required to reach the equivalence point? WebAssign will check your answer for the correct number of significant figures. 52.2 mL (this one is right)
(b) What is the pH at the equivalence point? WebAssign will check your answer for the correct number of significant figures.
(c) What is the pH of the solution after the addition of 18.6 mL of acid? WebAssign will check your answer for the correct number of significant figures.
(d) What is the pH of the solution after the addition of 86.0 mL of acid?
Explanation / Answer
a) let the volume of HCl required be x.so,
x*0.074 = 33*0.117
or x=52.1757 mL
b)the reaction is between a weak base and a strong acid. Hence, the pH at the equivalence point will be given by the formula
pH= 7 - 0.5pKb - 0.5*log(C)
where C is the concentration of the salt formed.
therefore, C=33*0.117/(33+52.1757)
=0.045
so pH = 7-0.5*4.76 - 0.5*log(0.045)
=5.29
c) a buffer will be formed after the addition of the 18.6 mL of the acid.
hence,
pH = 14- (pKb + log(salt/base))
=14 - (4.76 + log((18.6*0.074)/(33*0.117 - 18.6*0.074))
=9.496
d) the entire base will be used up.so,
[H+]=((86-52.5)*0.074/86)
=0.0288
so pH=-log(0.0288)
=1.54