Question

I did a lab and it's asking for the theoretical pH with the addition of HCl. The procedure:measure out 5.0ml of 1.0ml acetic acid solution in a 100 ml graduated cylinder. dilute the 5.0 ml w/ water to a final volume 50.0 mL&stir.measure out 1.00ml of 1.0 M NaOH&add it to the acetic acid solution.stir the mixture.calculate the theoretical pH.

Answer 1

Given that;

5.0ml of 1.0 M acetic acid solution in a 100 ml graduated cylinder. dilute the 5.0 ml w/ water to a final volume 50.0 mL

First calculate the molarity of dilute solution:

M1V1 = M2V2

1.0*5.0=M2*50

M= 0.1

Now number of moles of acetic acid:

1. M* 50 mL = 0.1 mole/ L* 0.05 L= 5*10^-3 Moles

CH3COOH + NaOH --à CH3COONa + H2O

Now calculate the number of moles of NaOH:

1.00ml * 1.0 M NaOH = 1.0 *10^-3 L*1.0 M NaOH

=1.0 *10^-3 L*1.0 moles NaOH

Here 1.0 *10^-3 L*1.0 moles of CH3COOH reacts with 1.0 *10^-3 L*1.0 moles NaOH so

4.0 *10^-3 L*1.0 moles CH3COOH remain in solution.

Now molarity of this acetic acid :

4.0 *10^-3 L*1.0 moles CH3COOH / total volume

4.0 *10^-3 L*1.0 moles CH3COOH / 50.0 +1.0 ml

4.0 *10^-3 L*1.0 moles CH3COOH / 0.051 L

=0.0784 M

: Ka = 10 ^ - 4.76=1.74 x 10^-5

CH3COOH <-----> CH3COO- + H+
initial concentration
0.0784
change
-x. . . . . . . . . . . . . . +x .. . . . +x
at equilibrium
0.0784-x .. . . . . . . . . .x .. . . . . .x

Ka = 1.74 x 10^-5 = [CH3COO-][H+]/ [CH3COOH]

1.74 x 10^-5 = (x)(x)/ 0.0784-x

x = [H+]= 1.168*10^-3 M

pH =- log [H+]= - log 1.168*10^-3 =2.93