AP Chemistry
WS 11.5 Key
.0008
(a) 𝑝𝐻 = − log(6.2 × 10−10 ) + log(
)
.0042
= 8.49
stoich table
1.
OH.0008
-.0008
0
HCN
.005
-.0008
.0042
CN0
+.0008
.0008
H2O
(b)
stoich table
OHHCN
CNH2O
.005
.005
0
-.005
-.005
+.005
0
0
.005
CN- is a conjugate base of a weak acid, therefore we complete a hydrolysis reaction. First determine the concentration
.005𝑚𝑜𝑙𝑒𝑠
of the CN- ion in solution. [𝐶𝑁 −] = .05𝐿+ .05𝐿 = 0.05𝑀
RICE table
CNH2O
HCN
OH.05
0
0
-x
+x
+x
.05-x
x
x
convert Ka to Kb
OH.0025
-.0025
0
𝐾𝑏 =
1𝑥10−14
6.2𝑥10−10
𝑥2
= 1.6𝑥10−5 =
= 8.94 × 10−4 = [𝑂𝐻−]
.05
𝑝𝑂𝐻 = − log(8.94 × 10−4 ) = 3.05
14 − 3.05 = 𝑝𝐻 = 10.95
(c) at half equivalence point
HCN
.005
-.0025
.0025
.0025
−10 )
𝑝𝐻 = − log(6.2 × 10
+ log(
)= 9.21
CN0
+.0025
.0025
H2O
.0025
OH.01
-.005
.005
(d) at 50 mL of NaOH. To calculate the moles of additional OH- added multiply total volume of base added by
concentration of the base. (total of 50mL of base used up to equivalence point and then an additional 50mL).
Also notice that after addition of base the moles of OH- are greater than CN- so we simply have a strong base
dissociation.
HCN
CNH2O
.005
0
-.005
+.005
0
.005
moles of base .005moles divided by total volume of 150 mL =
𝑝𝑂𝐻 = − log(. 033) = 1.48
14 − 1.48 = 12.51
2.
.005𝑚𝑜𝑙𝑒𝑠
.15 𝐿
= .033𝑀
3.