Answer to Question #182007 in General Chemistry for dasd

Question

Given:

Titrant: HCl

Analyte: NH3

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

pH = - log [H₃O⁺].

pH at HALF-WAY = pKab

pOH = -log 2.79x10^-6 = 5.55

pH = 14 - 5.55 = 8.45