If only 15.00 mL of 0.100 M HCI was titrated with 0.100 M NaOH, what differences

Question

If only 15.00 mL of 0.100 M HCI was titrated with 0.100 M NaOH, what differences would you expect in the resulting titration curve, as compared to the HCI-NaOH curve ptotted for this experiment? What similarities would there be between the two curves? 3. A student following the procedure in this lab titrated 25.00 mL of 0.103 M acetic acid, Ka = 1.8 times 10^-5, with 0.100M NaOH. Calculate the pH when the following volumes of NaOH were added: 0.00 mL NaOH 10.00 mL NaOH 28.75 mL NaOH 40.00 mL NaOH Show your work on additional paper. What volume of NaOH is needed to reach the equivalence point? What is the pH at the equivalent point? Show your work on additional paper.

Explanation / Answer

2) The resulting titration curve will not undergo any changes because concentration of both the acid and base are remaining the same that is why it will not same.

3)

Ka = [H+] / [Ac-]

Ka = [X][X] / [0.103M]

X=[H+]=(1.8x10-5 * 0.103)^1/2

pH=-log[H+]=-log 0.00136M

pH=2.86

Moles HAc=25.0mL x 0.103M=2.575 mmol

Moles OH-=10mL x 0.100M=1mmol

HAc +

è Ac-

mmol

1 mmol

------

+1 mmol

1.575mmol

using the Henderson-Hasselbalch equation.

pKa=-log Ka=-log 1.8x10-5

pKa=4.74

pH=4.54

Similarly you can do with rest other data.

HAc +

OH-

è Ac-

+H2O 2.575

mmol

1 mmol

------

C -1 mmol -1mmol

+1 mmol

E 1.575mmol 0

1.575mmol

using the Henderson-Hasselbalch equation.

pH=pKa+log [Base] [Acid]

pKa=-log Ka=-log 1.8x10-5

pKa=4.74

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.