Page 21 - Referance Guide For Pharmaceutical Calculations (Naplex, Fpgee and Ptce)

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-4
= 31.41 gm/mole Since [0.2- (1.2 x 10 )] is very small to calculate and
can be considered equal to 0.2.
223.(c) 0.331
-8
-4 2
= (1.2 x 10 ) = 7.2 x 10
Molarity is defined as the number of moles in 0.2
1000cc.
226.(b) The ionization constant Kb for Morphine
-6
-4
Number of moles = 50 = 0.331 moles base is 3 x 10 and C is 3 x 10 :
151
-6
Kb = 3 x 10
0.331 moles of ferrous sulfate are present in 1000cc C = 3 x 10
-4
-
therefore the molarity of the solution is 0.331 M. OH = ?

224.(c) 0.29% = −2
−
( − )

Moles of FeSO4 = 25 = 0.164 moles -
151.9 Since [C- OH ] is very small to calculate and can be
considered equal to C.
Moles of H2O = 1000 = 55.55 moles
-
−6
−4
18 OH = √ = √3 10 3 10

-5
-
Mole fraction of FeSO4 = 0.164 OH = 3 x 10 mole/liter
0.164 + 55.55
227.(c) The new hydroxyl ion concentration can be
= 0.0029 calculated by the following formula:

-
+
% Mole fraction of FeSO4 = 0.0029 x 100 Kw = [H3O ] x [OH ]
= 0.29%
-
+
OH = Kw/H3O
-
-14
-1
Mole fraction of H2O = 55.55 OH = 1 x 10 /1 x 10
-13
-
0.164 + 55.55 OH = 1 x 10

-8
= 0.997 228.(c) 0.66 x 10

% Mole fraction of H2O = 0.997 x 100 Kw = Ka x Kb
= 99.71% Ka = Kw/Kb

-14
225.(c) A 0.2 M solution of acetic acid is dissociated = 1 x 10
-6
-4
into 1.2 x 10 moles of hydrogen and acetate: 1.5 x 10

-8
-4
X = 1.2 x 10 moles = 0.66 x 10
C = 0.2 moles
229.(b) The solubility product of the salt (strong
2 electrolytes) can be calculated using following
= ( − ) formula:

(1.2 x 10 −4 2 + -
)
= Ksp = (Ag ) x (Cl )

(0.2−1.2 x 10 −4 ) = (1 x 10 ) x (1 x 10 )
-4
-4

-8
= 1 x 10
70