When the titration of HCN with NaOH is:
HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)
So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1
we need to get number of mmol of HCN = molarity * volume
= 0.2 mmol / mL* 10 mL = 2 mmol
so the number of mmol of NaOH = 2 mmol according to the molar ratio
so, the volume of NaOH = moles/molarity
= 2 mmol / 0.0998mL
= 20 mL
and according to the molar ratio so, moles of CN- = 2 mmol
∴the molarity of CN- = moles / total volume
= 2 mmol / (10mL + 20mL ) = 0.0662 M
when we have the value of PKa = 9.31 and we need to get Pkb
so, Pkb= 14 - Pka
= 14 - 9.31 = 4.69
when Pkb = -㏒Kb
4.69 = -㏒ Kb
∴ Kb = 2 x 10^-5
and when the dissociation reaction of CN- is:
CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq)
by using the ICE table:
∴ the initials concentration are:
[CN-] = 0.0662 M
and [HCN] = [OH]- = 0 M
and the equilibrium concentrations are:
[CN-] = (0.0662- X)
[HCN] = [OH-]= X
when Kb expression = [HCN][OH-] /[CN-]
by substitution:
2 x 10^-5 = X^2 / (0.0662 - X)
X = 0.00114
∴[OH-] = X = 0.00114
when POH = -㏒[OH]
= -㏒ 0.00114
POH = 2.94
∴PH = 14 - 2.94 = 11.06
Answer:
to which cations from the salt bridge migrate
Explanation:
A voltaic cell is an electrochemical cell that uses spontaneous redox reactions to generate electricity. It's composed of a cathode, an anode, and a salt bridge.
In cathode, the substance is gaining electrons, so it's reducing, in the anode, the substance is losing electrons, so it's oxidating. The flow of electrons is from the anode to the cathode.
The salt bridge is a bond between the cathode and the anode. When the redox reaction takes place, the substances produce its ions, so the solution is no more neutral. The salt bridge allows the solutions to become neutral and the redox reaction continues.
So, the cathode produces anions, which goes to the anode, and the anode produces cations, which goes to the cathode. Then, the cathode n a voltaic cell is the electrode to which cations from salt bridge migrate and where the reduction takes place.
I think you means the KO2 reacts with H2O. The equation of this reaction is 4KO2+2H2O->4KOH +3O2. The ratio of mole number of O2 and KO2 is 3:4. So the mole number of O2 produced is 0.500/4*3=0.375 mol.
