Consider the titration of 52.0 mL of 1.2 M NaOH with 1.0 M HCl. Find the pH at a
ID: 791061 • Letter: C
Question
Consider the titration of 52.0 mL of 1.2 M NaOH with 1.0 M HCl. Find the pH at all the following points.
A). At the start of the titration. I know this answer is 14.08
B). At the equivalence point.
C). After the addition of a large excess of acid (in comparison with the acid volume needed to reach the equivalence point).
D). Tell how many milliliters of acid are required to reach the equivalence point.
I have no idea where to start for B, C and D. Please show all steps and correct answers so I can understand.
Thank you!!
Explanation / Answer
a)pH=14-pOH
=14+log(1.2)
=14.08
b)at equivalence point, there is no acid or base.
hence pH=7
c)excess of acid .so,
[H+]=1
so pH=-log(1)
=0
d)let the volume be v.so,
NaOH+ HCl---> NaCl+ H2O
equating the number of moles,
M1V1=M2V2
or 1.2*52=1*v
or v=62.4 mL