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Concentration of HCl solution= 0.1N

Mass of Ag deposited in coulometer = 0.1209 g

Movement of boundary = 7.50 cm

Cross-section of tube= 1.24 $c{{m}^{2}}$

Write an answer to the nearest integer after multiplying the transport number calculated with 10.

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${{H}^{+}}$ ions = $\dfrac{l\times a\times c}{1000Q}$

- We are being provided with the information that Concentration of HCl solution= 0.1N,

Mass of Ag deposited in coulometer = 0.1209 g, Movement of boundary = 7.50 cm, and

Cross-section of tube= 1.24 $c{{m}^{2}}$.

- We will calculate the transport number of ${{H}^{+}}$ ions by moving the boundary method. We will calculate this by the formula:

${{H}^{+}}$ ions =$\dfrac{l\times a\times c}{1000Q}$

Where l is the length of migration that is equal to 7.50 cm. a is the area of cross – section is equal to 1.24 $c{{m}^{2}}$. And c is the concentration of solution equal to 0.1 N.

- As we know that Q is the charge passed which is equal to the equivalent of Ag deposited.

We can write the formula for this as: $Q =\dfrac{W}{E}$

- Now, by putting all the values given in this equation we get:

$\begin{align}

& =\dfrac{0.1209}{108}Faraday \\

& \therefore {{t}_{{{H}^{+}}}} =\dfrac{7.5\times 1.24\times 0.1\times 108}{0.1209\times 1000} \\

& = 0.8308 \\

\end{align}$

- Hence, we get the value of the transport number of ${{H}^{+}}$ ions as 0.8308.

Therefore, we get the correct as 10 X (0.8308) = 8.308 or 8

- Now, further as we know that ${{t}_{{{H}^{+}}}}+{{t}_{C{{l}^{-}}}}=1$

${{t}_{C{{l}^{-}}}}$ = 1 - 0.8308 = 0.1692

transport no. of $C{{l}^{-}}$ ion is 0.1692.

- Hence, we can conclude that the value of the transport number of ${{H}^{+}}$ ions is 0.8308.