Sodium cyanide, nacn, is a salt formed from the neutralization of the weak acid hydrocyanic acid, hcn, with the strong base sodium hydroxide. given that the value of ka for hydrocyanic acid is 4.90×10−10, what is the ph of a 0.265 m solution of sodium cyanide at 25∘c?

pH = 11.37

Explanation:

Sodium cyanide will dissociate into sodium ion and cyanide ion. This cyanide ion will get hydrolyzed.The ICE table for hydrolysis of cyanide ion is:

Initial                      0.265                                  0         0

Change                 -x                                           +x        +x

Equilibrium        0.265-x                                     x          x

This K is Kb of HCN

Kb = Kw / Ka

Putting values

x can be ignored in denominator as Kb is very low

x= 2.33X10⁻³ M = [OH⁻]

pOH = -log[OH⁻]

pOH = -log(2.33X10⁻³ )

pOH = 2.63

pH = 14- pOH = 14-2.63 = 11.37  

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

that's what i got you can count the number of atoms