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
Answer:
0.444 mol/L
Explanation:
First step is to find the number of moles of oxalic acid.
n(oxalic acid) = 
Now use the molar ratio to find how many moles of NaOH would be required to neutralize 
of oxalic acid.
n(oxalic acid): n(potassium hydroxide)
1 : 2 (we get this from the balanced equation)
: x
x = 0.0111 mol
Now to calculate what concentration of KOH that would be in 25 mL of water:
(If this is correct, can I have Brainlist?)
Answer:
D) anomalous volcanoes such as those in Hawaii
Answer:
CaCl2 + H2O Ca2+(aq) + 2Cl-(aq)
Explanation:
CaCl2 + H2O Ca2+(aq) + 2Cl-(aq)
When CaCl2 is dissolved in H2O (water) it will dissociate (dissolve) into Ca+2 and Cl- ions.
The dissolution of calcium chloride is an exothermic process.
It’s c I just did this so yea c
