Answer to Question #183178 in General Chemistry for sdafds

Question

8.0 mL of 0.10 M HC2H3O2 is titrated with 0.10 M NaOH. Perform the following calculations & Show ALL work

1. pH of the acid before any NaOH is added
2. Volume of 0.10 M NaOH needed to reach the equivalence point
3. pH at the equivalence point
4. Volume of 0.10 M NaOH needed to reach the half-way point
5. pH at the half-way point

Answer 1

$PH=- log [H^+]$

$PH=-log[0.10]=1$

NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

Volume of NaOH = 8.0 ml

pH of half-way point.

8 mL of 0.1 M acetic acid = 0.1 M / 1000 mL x 8.0 mL = 8 x 10^-4 moles.

At the half-way point, 1/2 the original number of moles has been titrated.

As the molarity of the NaOH is also 0.1 M, this means 4.0 mL of NaOH has been added. Also, only 4 x 10^-4 moles if acetic acid remains and 4 x 10^-4 acetate has been produced.

Concentration of acetic acid = 4 x 10^-4 moles / 12 mL x 1000 mL/L =

3.33 x 10^-2M. As it is the half-way point, the acetate concentration is also 3.33 x 10^-2 M.

According to the Henderson-Hasselbalch equation,

pH = pKa + log [acetate] /[acetic acid] = pKa + log 1 = pKa.

In this case the pH is equal to the pKa which is 4.75