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

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

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 + ]

p.153: #84 SO 3 -2 + H 2 O HSO 3 - + OH - I#M - 0 0 C - x - +x +x E(#M - x) - x x SO 3 -2 + H 2 O HSO 3 - + OH - I#M - 0 0 C - x - +x +x E(#M - x) - x x

pH = 9.69 pOH = 4.31 [OH - ] = 4.8978 x 10 -5 M So x = 4.90 x 10 -5 M pH = 9.69 pOH = 4.31 [OH - ] = 4.8978 x 10 -5 M So x = 4.90 x 10 -5 M

Kb = Kw/Ka = 1.0 x 10 -14 / 1.0 x 10 -7 = 1.0 x 10 -7 Kb = [HSO 3 - ][OH - ]/ [SO 3 -2 ] = (1.0 x 10 -7 ) 2 / (#M - 4.90 x 10 - 5 ) *Assume x (4.90 x 10 -5 ) is so small so… 1.0 x 10 -7 = (1.0 x 10 -7 ) 2 / (#M) [SO 3 -2 ] = 0.024M Kb = Kw/Ka = 1.0 x 10 -14 / 1.0 x 10 -7 = 1.0 x 10 -7 Kb = [HSO 3 - ][OH - ]/ [SO 3 -2 ] = (1.0 x 10 -7 ) 2 / (#M - 4.90 x 10 - 5 ) *Assume x (4.90 x 10 -5 ) is so small so… 1.0 x 10 -7 = (1.0 x 10 -7 ) 2 / (#M) [SO 3 -2 ] = 0.024M

IV. 16. TITRATIONS (Hebden p. 154-158)

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

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

Example: If 25.9 mL of H 3 PO 4 with an unknown molarity react with 34.6 mL of 0.400 M KOH according to the rx, what is the molarity of the H 3 PO 4 ?

H 3 PO 4 + KOH K 3 PO 4 + H 2 O 25.9 ml 34.6 ml # M 0.400 M 25.9 ml 34.6 ml # M 0.400 M

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

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

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