Acids - Base Titration. 20.0 ml acetic acid is titrated to the endpoint with 32.

Question

Acids - Base Titration. 20.0 ml acetic acid is titrated to the endpoint with 32.25 ml 0.125 m NaOH. Define endpoint and equivalence point. Distinguish between weak acids and strong acids. Is acetic acid a weak, acid or a strong acid? Give the balanced molecular, total, and net ionic equations for the reaction between acetic acid and NaOH. Molecular: Total: Net: What is the morality of the acetic acid? Equal volumes of 0.250 M acetic acid and water are combined; a 50.0 mL portion of this solution is titrated with 0.100 M NaOH. What volume of NaOH is required to reach the endpoint?

Explanation / Answer

4a.

The equivalence point is a special point in the titration, it is the point in which we have neutralized all the acid and base

that is

mol of acid = mol of base

The endpoint is given when we stop the titration. Since it is pretty hard to exactly end in the equivalence point, the endpoint is typically larger than that of the equivalence point.

In this case

equivalence point = endpoint, since no data is given

4b.

weak acid = any specie that will act as an acid, that is, will donate protons ( H+ ions ) but will not completely dissociate.

ex. Acetic acid; HA <--> H+ and A-

strong acid = any specie that will act as an acid, that is, will donate protons ( H+ ions ) AND WILL completely dissociate.

ex. Hydrochloric acid; HCl --> H+ and Cl-

c.

Balanced Molecular Equation

CH3COOH(aq) + NaOH(aq) ---> NaCH3COO(aq) + H2O(l)

Total Equation

CH3COO-(aq) + H+(aq) + Na+(aq) + OH-(aq) ---> Na+(aq) + CH3COO-(aq) + H2O(l)

Net Ionic (get rid of spectator ions marked in bold)

CH3COO-(aq) + H+(aq) + Na+(aq) + OH-(aq) ---> Na+(aq) + CH3COO-(aq) + H2O(l)

H+(aq) + OH-(aq) ---> H2O(l)

d.

For Molarity of Acid,

assume the acid's mol = base's mol in th eequivalence point

so

mol of acid = mol of base

mol of base = [base]*Vbase

mol of base = [base]*Vbase = 0.125*32.25 = 4.03125 mmol of base (note we use milimol)

mol of acid = mol of base = 4.03125 mmol of acid

but we need [acid]

sop

[acid] = mmol of acid / mL

[acid] = 4.03125 / 20 = 0.2015625 M

[acid] = 0.2015625 M

e.

If equal volumes then

50 mL = V1 + V2

V1 = V2

so

2*V = 50 mL

V = 25 mL of each species

mol of acid = M*V = 0.25*25 = 6.25 mmol of acid (since we used mL)

then

mmol of base = mmol of acid

mmol of base = 6.25 mmol of base needed

mmol = M*V

V = mmol/M

V = 6.25 / 0.125 = 50 mL

Vbase = 50 mL

we need 50 mL of base for this neutralizaiton

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