CHEM1002 2008-N-5 November 2008
โข Calculate the pH of a 0.020 M solution of Ba(OH)2.
Marks
1
Ba(OH)2 is a strong base so it will completely dissociate in solution:
Ba(OH)2(s) ร ๏ Ba2+(aq) + 2OH-(aq)
As each Ba(OH)2 dissociates to make 2OH-, a 0.020 M solution has [OH-(aq)] =
0.040 M.
By definition, pOH = -log10[OH-(aq)] so pOH = -log10(0.040) = 1.40.
As pH + pOH = 14.00,
pH = 14.00 โ 1.40 = 12.60
pH = 12.60
โข Calculate the pH of a 0.150 M solution of HNO2. The pKa of HNO2 is 3.15.
3
As HNO2 is a weak acid, [H3O+] must be calculated using a reaction
table:
HNO2
H2O
H3O+
NO2-
initial
0.150
large
0
0
change
-x
negligible
+x
+x
final
0.150 โx
large
x
x
The equilibrium constant Ka is given by:
Ka =
๐๐๐![๐๐๐
!]
[๐๐๐๐] =
๐๐
๐.๐๐๐!๐
As pKa = -log10Ka, Ka = 10โ3.15 and is very small, 0.150 โ x ~ 0.150 and hence:
x2 = 0.150 ร 10โ3.15 or x = 1.03 ร 10-2 M = [H3O+]
Hence, the pH is given by:
pH = โlog10[H3O+] = โlog10(1.03 ร 10-2) = 1.987
pH = 1.987
ANSWER CONTINUES ON THE NEXT PAGE
