According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
The generic equation for a reaction between an acid and water is

When an acid "reacts" with water, water acts as the base that accepts the proton (H+) from the acid. The remaining ion that is formed after the acid has donated its proton is called the conjugate base (

), and the conjugate acid-base pair is

-

.
Hydrogen sulfate (

) is an ion from sulfuric acid. It is still an acid in itself and can "react" with water ((

) to form the sulfate (

) and hydronium (

)ions.

Based on the previous discussion,

is identified to be the conjugate of the acid

.
Thus, the conjugate acid-base pair is
.
Answer:
hello your question is incomplete the options are missing
Determine the resultant velocity for the plane when it is travelling
i) To the east
ii) To the west
answer :i) 270 i
ii) -150 i
Explanation:
velocity of Airplane = 210 m/s
wind velocity = 60 m/s to the east
The resultant velocity for the plane when it is travelling
let the velocity of the wind = V2
velocity of the plane = v1
i) The resultant velocity for the plane when travelling to the east
Vr = V2 i + V1 i
Vr= 60i + 210i = 270i
ii) resultant velocity when the plane is travelling to the west
Vr = - V1 i + V2i
= -210i + 60 i = -150 i