Question

Consider the titration of a 25.0 mL sample of 0.100 M CH3COOH (Ka= 1.8e-5) with 0.130 M NaOH. Determine the pH after adding 4.0 mL of base beyond the equivalence point.

Answer 1

Ans :- The titration reaction can be written as

CH₃COOH + H₂O CH₃COO^- + H₃O⁺

We know , 1ml = 0.001L

               25 ml = 0.025L

initial moles of CH₃COOH = 0.025 L x 0.100 mol/L

                                       = 0.0025 mol CH₃COOH

At equivalence point,

moles of NaOH added = initial moles of CH₃COOH

Therefore, moles of NaOH added = 0.0025 mol

Now, Volume of NaOH added to reach equivalent point = 0.0025 mol NaOH / 0.130 mol/L NaOH

                                                                                = 0.01923L

Now , 1L = 1000ml

0.01923 L = 19.23ml

Therefore, Volume of NaOH added to reach the equivalent point = 19.23 ml

Now, after adding 4.00 ml of base NaOH beyond the equivalent point, the amount of excess NaOH can be calculated as follows;-

Number of moles of excess OH^- = 0.00400L x 0.130 mol/L

                                               = 0.00052 mol OH^-

total Volume = Initial volume of CH₃COOH + Volume of NaOH added to reach the equivalent point + Volume of

                      NaOH added after equivalence point

                   = 25.00 + 19.23+ 4.00 ml

   = 48.23 ml

We Know 1ml=0.001L

   48.23 ml = 0.04823L

therefore,

total volume = 0.04823L

[OH^-] = Number of moles of OH^- / total volume

   = 0.00052 mol / 0.04823L

   = 0.01078 M

Now ,

[H₃O⁺] [OH^-]=K_w

Where K_w = ionic product of water

   = 1x10^-14

[H₃O⁺]= K_w / [OH^-]

= 1x10^-14 / 0.01078

   =9.27 x 10^-13 M

Now,

pH = -log[H₃O⁺]

   = -log(9.27 x 10^-13)

   = 12.032