When a strong base is gradually added dropwise to a weak acid, the pH changes at

Question

When a strong base is gradually added dropwise to a weak acid, the pH changes at each addition. When the appropriate quantity base been added to react with all of the acid, the pH changes sharply, indicating the endpoint of the titration. A plot of pH versus volume of base added gives what is known as a titration curve. Consider the titration of 25.00 mL, of 0.1000 M acetic acid with 0.1000 M NaOH. a. Write the equation for the titration reaction. b. Determine the volume of NaOH solution required to reach the endpoint. 25.00 ml c. The chart below has entries for several steps along the titration curve. To calculate the pH at each step, you first must understand the chemistry at that step, and then you can decide the appropriate method to calculate the pH. For each volume listed use the appropriate equation to calculate the solution [H_3O^+] and pH. Show all your work.

Explanation / Answer

Q2.

a

the reaction

acetic acid = CH3COOH and base, NaOH

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

b)

the ratio is 1:1

so

mmol of acid = MV = 25*0.1 = 2.5 mmol of acetic acid

mmol of base = 2.5 mmol of NAOH

volume required = 2.5/0.1 25 mL of bases

c)

pH = -log(H+)

so...

V = 0 ml --> just H+ is present, use

Ka = [H+][A-]/[HA]

V = 5 mL, 12.5 mL, 20 mL --> use buffer equation --> pH = pKa + log(A-/HA)

calcualte

mmol of A- = 0 + Mbase*Vbase = 0.1*(Vbase)

mmol of HA left = 2.5 - 0.1*(Vbase)

pKa = 4.75 always

finally

V = 25 mL

this is equivalence point

CH3COO- the acetate will hydrolise

CH3COO- + H2O <-> CH3COOH + OH-

Kb = [CH3COOH ][OH-]/[CH3COO-]

calculate OH- then pOH then pH

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