Answer : The final concentration of [tex]H_2SO_4[/tex] is, 5.47 M
Explanation :
First we have to calculate the charge.
Charge = Current × Time
Charge = 2.50 A × 7.50 hr
Charge = 2.50 A × 7.50 × 3600 s
Charge = 67500 C
Now we have to calculate the moles of electrons.
[tex]\text{Moles of electrons}=\frac{Charge}{\text{Faraday constant}}[/tex]
[tex]\text{Moles of electrons}=\frac{67500}{96500}[/tex]
[tex]\text{Moles of electrons}=0.699mol[/tex]
That means,
Moles of [tex]H_2SO_4[/tex] consumed = Moles of electrons = 0.699 mol
Now we have to calculate the initial moles of [tex]H_2SO_4[/tex]
[tex]\text{Initial moles of }H_2SO_4=Concentration\times Volume[/tex]
[tex]\text{Initial moles of }H_2SO_4=5.00M\times 1.50L=7.5mol[/tex]
Final moles of [tex]H_2SO_4[/tex] = 7.5 + 0.699 = 8.199 mol
Now we have to calculate the final concentration of [tex]H_2SO_4[/tex]
[tex]\text{Concentration of }H_2SO_4=\frac{Moles}{Volume}[/tex]
[tex]\text{Concentration of }H_2SO_4=\frac{8.199mol}{1.50L}=5.466M\approx 5.47M[/tex]
Therefore, the final concentration of [tex]H_2SO_4[/tex] is, 5.47 M
The final concentration of the solution is 4.534 M.
Using Faraday's first law of electrolysis;
m = ZIt
m = mass of substance
Z = electrochemical equivalent
I = current
t = time taken
Also;
Z = Equivalent weight/ Faraday's constant
Z = 2.5 × 7.5 × 60 × 60/96500 = 0.699
Number of moles of H^+ = 0.699/2 = 0.35 moles
Molarity of H^+ = 0.35 moles/ 1.50 L = 0.233 M
Final concentration = 5. 00 M - 2( 0.233 M) = 4.534 M
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