Question

calculate the theoretical ph after the first addition of acid (HCL) for 10ml of .2m acetic acid and 10ml .2m sodium acetate. first addition of acid is 1ml of      .099233m HCL.

Answer 1

We have 10 mL of 0.2 M Acetic Acid (AH) and 10mL of 0.2M Sodium Acetate(A-). The mixture total volume is 20mL(0.02L).

Therefore,

The no. of moles of AH = 0.2Mx0.010L = 0.002 moles

The no. of moles of A- = 0.2Mx0.010L = 0.002 moles

After the first addition of 1mL of 0.09923 M HCl, the A- reacts with HCl to form AH.

No. moles of HCl added = 0.099233 M x 0.001 L = 0.0001 moles

Therefore, after the addition;

The no. of moles of AH = 0.002 moles + 0.0001 moles = 0.0021 moles

The no. of moles of A- = 0.002 moles -0.0001 moles = 0.0019 moles

Total volume = 10mL +10mL + 1 mL = 21mL = 0.021 L

The concentration [AH] = no. moles/volume in L = 0.0021 moles/0.021 = 0.1 M

The concentration [A-] = no. moles/volume in L = 0.0019 moles/0.021 = 0.09M

Henderson–Hasselbalch equation:

pH = pKa + log [A-]/[AH]

pH = 4.75 + log (0.1M/0.09M) = 4.795 = 4.8

Therefore, the pH of the solution = 4.8