Answer to Question #183179 in General Chemistry for sad

Question

Given:

Titrant: HCl

Analyte: NH3

Answer all of these with workout for each please!

Find-> pH at HALF-WAY point

Find--> Volume at HALF-WAY point

Find--> pH at EQUIVALENCE POINT

Find--> Volume at EQUIVALENCE POINT

Data

Titration Lab

HCl + NH3

Volume of 0.10 M NH3 added (mL)--> 8.05 mL

interval=0.5 mL

Volume 0.10 M HCl Added (mL)

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

4.50

5.00

5.75

6.00

6.50

7.00

7.50

8.00

9.00

PH

10.05

10.16

10.00

9.87

9.76

9.61

9.53

9.40

9.31

9.20

9.10

8.91

8.74

8.30

7.60

6.79

3.55

3.18

Answer 1

Part (a): HC₇H₅O₂ + OH^- ===> C₇H₅O₂^- + H2O

Instead of typing HC₇H₅O₂ for the acid, I will simply call it HA. We then have

HA + OH^- ==> A^- + H2O

Initial moles HA = 100.0 ml x 1 L/1000 ml x 0.10 mol/L = 0.01 moles HA

Initial moles OH- to reach 1/2 equivalence = 1/2 x 0.01 moles = 0.005 moles (= 50 ml of 0.1 M NaOH)

At this point of 1/2 equivalence we now have 0.005 moles HA and 0.005 moles A-. Looking at an ICE table...

HA + OH- ===> A- + H2O

0.01....0.005............0..........0.........Initial

-0.005...-0.005......+0.005.............Change

0.005.......0.............0.005..............Equilibrium

Final volume = 100 ml HA + 50 ml OH- = 150 ml = 0.150 L

Final [HA] = 0.005 mol/0.15 L = 0.0333 M

Final [A-] = 0.005 mol/0.15 L = 0.0333 M

Using the Henderson Hasselbalch equation (where pKa = -log Ka = -log 6.4x10^-5 = 4.19)

pH = pKa + log [A-]/[HA] = 4.19 + log [0.0333]/[0.0333] = 4.19 + 0

pH = 4.19

At full equivalence point, all the HA is neutralized (none left) and it has all been converted to A-. In this case that will be 0.01 moles HA ==> 0.01 moles A^- in a final volume of 200 ml (0.2 L). The 0.2 L comes from 100 ml of HA + 100 ml of NaOH.

Final [A-] = 0.01 mol/0.2 L = 0.05 M

Now, we look at the hydrolysis of A^- as follows:

A^- + H2O ==> HA + OH^- (note that A- acts as a base in this case, so we need to use Kb)

KaKb = Kw = 1x10^-14

Kb = 1x10^-14/6.4x10^-5

Kb = 1.56x10^-10

Kb = 1.56x10^-10 = [HA][OH^-]/[HA] = (x)(x)/0.05

x² = 7.8x10^-12

x = 2.79x10^-6 = [OH^-]

pOH = -log 2.79x10^-6 = 5.55

pH = 14 - 5.55 = 8.45