6
Given,
In a particular Redox reaction Fe^{2+} is oxidized to Fe^{3+} and Au^{3+} is reduced to Au
Number of electrons are transferred in this reaction = ?
First we have to write the complete to Redox reaction:
In a particular Redox reaction Fe^{2+} is oxidized to Fe^{3+} and Au^{3+} is reduced to Au.
At anode: oxidation (loss of electrons)
Fe^{2+} ➝ Fe^{3+} + e^{-}
Multiplying the above equation by 3, we get
3 × [Fe^{2+} ➝ Fe^{3+} + e^{-}]
3Fe^{2+} ➝ 3Fe^{3+} + 3e^{-} ----------(I)
At cathode: reduction (gain of electrons)
Au^{3+} + 3e^{-} ➝ Au -------------(II)
Now we have to write the completed redox reaction
Adding equation (I) and (II),
3Fe^{2+} ➝ 3Fe^{3+} + 3e^{-}
Au^{3+} + 3e^{-} ➝ Au
Adding both the above equation, we get
3Fe^{2+} + Au^{3+} ➝ 3Fe^{3+} + Au
So, 3 moles of electrons are transferred in this reaction
37)
Calculating the pH of a buffer solution = ?
Given,
[HC_{2}H_{3}O_{2}] = 0.317 M
[NaC_{2}H_{3}O_{2}] = 0.171 M
K_{a} for HC_{2}H_{3}O_{2} = 1.81 × 10^{-5}
First we have to calculate pK_{a} for HC_{2}H_{3}O_{2}:
K_{a} for HC_{2}H_{3}O_{2} = 1.81 × 10^{-5}
The formula used here is
pK_{a} = -logK_{a}
pK_{a} = -log(1.81 × 10^{-5})
pK_{a} = 4.742
Now we can calculate the pH of a buffer solution:
pH = pK_{a} + log[salt]/[acid]
pH = pK_{a} + log[NaC_{2}H_{3}O_{2}]/[HC_{2}H_{3}O_{2}]
pH = 4.742 + log(0.171 M/0.317 M)
pH = 4.742 + (-0.268)
pH = 4.474
pH ≈ 4.47
(Rounding answer to 2 decimal places)
So, pH of a buffer solution is 4.47
Conclusion,
36) number of electrons transferred are = 6 moles
37) The pH of a buffer is 4.47 .