Download presentation

Presentation is loading. Please wait.

Published byRyan Davies Modified over 2 years ago

1
IV.15. CALCULATIONS INVOLVING Kb (Hebden p. 152 – 154)

2
Need to calculate Kb from Ka table. Basic solutions, so use [OH - ], not [H 3 O + ] Need to calculate Kb from Ka table. Basic solutions, so use [OH - ], not [H 3 O + ]

3
p.153: #84 SO H 2 O HSO OH - I#M C - x - +x +x E(#M - x) - x x SO H 2 O HSO OH - I#M C - x - +x +x E(#M - x) - x x

4
pH = 9.69 pOH = 4.31 [OH - ] = x M So x = 4.90 x M pH = 9.69 pOH = 4.31 [OH - ] = x M So x = 4.90 x M

5
Kb = Kw/Ka = 1.0 x / 1.0 x = 1.0 x Kb = [HSO 3 - ][OH - ]/ [SO 3 -2 ] = (1.0 x ) 2 / (#M x ) *Assume x (4.90 x ) is so small so… 1.0 x = (1.0 x ) 2 / (#M) [SO 3 -2 ] = 0.024M Kb = Kw/Ka = 1.0 x / 1.0 x = 1.0 x Kb = [HSO 3 - ][OH - ]/ [SO 3 -2 ] = (1.0 x ) 2 / (#M x ) *Assume x (4.90 x ) is so small so… 1.0 x = (1.0 x ) 2 / (#M) [SO 3 -2 ] = 0.024M

6
IV. 16. TITRATIONS (Hebden p )

7
Titration Process in which a measured amount of a solution is reacted with a known volume of another solution (one of the solutions has an unknown concentration) until a desired EQUIVALENCE POINT is reached

8
Equivalence Point The point in a titration where the ratio of the moles of each species involved exactly equals the ratio of the coefficients of the species in the balanced rx equation

9
Example: If 25.9 mL of H 3 PO 4 with an unknown molarity react with 34.6 mL of M KOH according to the rx, what is the molarity of the H 3 PO 4 ?

10
H 3 PO 4 + KOH K 3 PO 4 + H 2 O 25.9 ml 34.6 ml # M M 25.9 ml 34.6 ml # M M

11
% Yield = mass of product obtained x 100% mass of product expected

12
% Purity = mass of pure reactant x 100% mass of impure reactant

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google