Answer the following regarding the titration of 25.00 mL of a 0.070 M weak acid
ID: 1019705 • Letter: A
Question
Answer the following regarding the titration of 25.00 mL of a 0.070 M weak acid solution with 0.10 M NaOH.a. How many mL of 0.10 M NaOH are required to reach the first equivalence point?
b. How many total mL of 0.10 M NaOH are required to reach the second equivalence point?
c. What is the relationship between the volume of NaOH needed to reach the first and second equivalence points? Answer the following regarding the titration of 25.00 mL of a 0.070 M weak acid solution with 0.10 M NaOH.
a. How many mL of 0.10 M NaOH are required to reach the first equivalence point?
b. How many total mL of 0.10 M NaOH are required to reach the second equivalence point?
c. What is the relationship between the volume of NaOH needed to reach the first and second equivalence points? Answer the following regarding the titration of 25.00 mL of a 0.070 M weak acid solution with 0.10 M NaOH.
a. How many mL of 0.10 M NaOH are required to reach the first equivalence point?
b. How many total mL of 0.10 M NaOH are required to reach the second equivalence point?
c. What is the relationship between the volume of NaOH needed to reach the first and second equivalence points?
Explanation / Answer
a)
Write the first ionisation reaction as shown below
H2A + NaOH -----> HA- + H2O
moles of weak acid, H2A = 25.00 mL*0.070 M = 1.75*10^(-3)mol
At equivalent point moles of acid is equal to moles of base.
Volume of NaOH = moles of weak acid/0.10 M NaOH
= 1.75*10^(-3)mol / 0.10 M NaOH
= 17.5 mL
b)
Write the second ionisation reaction as shown below
HA- + NaOH -----> A2- + H2O
moles of weak acid, HA- = 1.75*10^(-3)mol
At equivalent point moles of acid is equal to moles of base.
Volume of NaOH = moles of weak acid/0.10 M NaOH
= 1.75*10^(-3)mol / 0.10 M NaOH
= 17.5 mL
Therefore, the volume of naoh at second equivalent point is double the volume at first equivalent point.
c) Total volume, V1 + V2
= 17.5 mL + 17.5 mL
= 35 mL
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