Consider the titration of 100mL of 0.30M acetic acid (CH 3 COOH, K a = 1.8 x 10

Question

Consider the titration of 100mL of 0.30M acetic acid (CH3COOH, Ka = 1.8 x 10-5) with 1.0M NaOH.

a) The titration starts as a solution of acetic acid, a weak acid. Write the reaction of CH3COOH with water and use an ICE table to calculate the pH of this solution.

b) As NaOH is added, some of the CH3COOH is neutralized, creating a buffer. Write the neutralization reaction between NaOH and CH3COOH and label the conjugate acid-base pair that forms the buffer.

c) Determine the equivalence point of the reaction.

d) Calculate the pH of the solution after addition of different volumes of NaOH.

0mL of 1.00M NaOH added 30mL of 1.00M NaOH added 15mL of 1.00M NaOH added 32mL of 1.00M NaOH added 20mL of 1.00M NaOH added 40mL of 1.00M NaOH added

Explanation / Answer

Answer: according to question the chemical equation is like

b] neutralysation reaction CH3COOH + NaOH< -----------------> CH3COONa + H2O

acid base conjugate base of acetic acid

here it is react with the ratio of 1:1

a} the reaction of CH3COOH with water is like

CH3COOH + H2O <--------> CH3COO- + H3O

c] the equevalence point is the point on which the number of moles of acid = number of moles base means the completely neutralysed .

hence number of moles of acetic acid for 100ml = 0.30 * 100/1000 = 0.03mol

means moles of acid = moles of base

we get volume around 30 ml it is the equivalent point.

D] Here in this part we have to use the henderson equation

Ph = PKa + log [salt]/[acid]

now concentration of acid means acetic acid is molarity for 100 ml solution = moles / volume in L

molarity = 0.03/ 0.1 = 0.3 M

and in this part the concentration of salt is changes hence now calculate the concentration of salt

like : 1] 15 ml of 1.00M NaOH

molarity = 1.00/ 0.015 = 66.67 M

And the value of PKa = -log[Ka] = -log[1.8* 10-5]= 4.745

now put the all values in Henderson equation we get

ph=4.745 + log[66.67]/[0.3] = 7.092

Hence by using this method we can calculate all values of Ph for different values of NaOH

I hope this explanation will help you .

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