Question

For the titration of 20.00 mL of 0.100 M Acetic Acid (CH3COOH, Ka = 1.8 � 10-5), calculate the pH after the addition of the following volumes of 0.100 M NaOH.

0.00mL

21.00mL

30.00 mL

Answer 1

Answer: Here the acetic acid is dissociated by using ICE table we get

CH3COOH <---------------> CH3COO - + H +

initial 0.100 0 0

0.100 -X X X   

Ka = X²/ 0.100 - X = 1.8 * 10^-5

here 0.100 -X is nearly = 0.100 so we get

X^{2 =} 0.18 * 10^-5   = 1.8 * 10^-6

X = 1.3416 * 10^-3

Hence the concentration of H + ion when 0 ml of NaOh is added

So the required PH = - log [ H+]

= - log [ 1.3416 * 10^-3]

= 2.87

Hence the PH at 0.0 ml of NaoH = 2.87

2) Now , the number of moles of 20.0 ml of Acetic acid = 0.100 * 20 / 1000 = 0.002 mol

and the number of moles of 21.0 ml of NaOH = 0.100 * 21 /1000 = 0.0021 mol

Hence we know that at the end point number of moles of acid = number of moles of base

and hence the nummber of moles of salt = 0.002 mol

Now using Henderson's equation we get .

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

Now PKa = -log [ ka ]

= -log [ 1.8 * 10^-5]

= 4.745

Now Ph = 4.745 + Log [ 0.002]/[0.002 ]

Ph = 4.745

Hence the requred Ph when 21 ml of NaoH is used is 4.745.

Note the third part is same as second part so just try to solve it out by using same procedure .

All the best .