Calculating the mole of electrons form the current

[bross7]

In the chlor-alkali industry, Cl₂(g) and NaOH(aq) are produced by electrolysis of an aqueous solution ofo sodium chloride. If a 1000L tank of NaCl(aq) at pH 7 is electrolysed for one hour with a current of 1.00x10⁴A, the final pH will be:

I am completely stuck on how to find the solution to the problem aside from calculating the mole of electrons form the current:

I = C/t
10000 = C/3600s
C = 3600000C

mol e^- = 360000000/96490Cmol^-1
mol e^- = 373.1

After that I am stuck.

 

[osskall]

Huh? You're stuck?
Well, what you want to know is the amount OH- produced, right?

How many electrons do you need to produce one OH- ion?
For this you should look at the reactions occurring at the anode and at the cathode.
At the anode, Cl- is converted to 1/2 Cl2(g) and an electron, at the cathode, water is reduced to hydrogen gas, leaving OH- behind. It is up to you to calculate how many. Hint: look only at the charged species in your cathode reaction and you'll see immediately how many OH- are produced for each electron? (oh, did I say too much? )

From this you get the pH... if you assume the volume of the water to be unchanged (see below)!

(Maybe we should check if there still are 1000L of water? We are after all consuming water and any decrease of the water volume would of course affect the pH - but given the question I guess we can assume the volume is unchanged. As an exercise, however, and to get an idea of the different proportions, I think you should calculate the amount of water consumed. Hint: how many oxygens are there in water? and in OH-? Given that water is reduced to OH- and H2, how many water molecules correspond to one OH- ion? What is the volume corresponding to this amount, given the density 1 g/ml and molar mass 18 g/mol?)

 

[wais]

To calculate the final pH, we need to consider the products of the electrolysis reaction. In the chlor-alkali industry, the electrolysis of sodium chloride results in the production of chlorine gas (Cl2) and sodium hydroxide (NaOH). The overall reaction can be represented as follows:

2NaCl(aq) + 2H2O(l) → Cl2(g) + 2NaOH(aq) + H2(g)

From this reaction, we can see that for every 2 moles of electrons transferred, 1 mole of chlorine gas and 2 moles of sodium hydroxide are produced. Therefore, the number of moles of electrons calculated earlier (373.1) can be used to determine the number of moles of chlorine gas and sodium hydroxide produced.

Moles of Cl2 = 373.1/2 = 186.55

Moles of NaOH = 2(186.55) = 373.1

To calculate the final pH, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid])

In this case, the base is NaOH and the acid is HCl (formed from the dissociation of H2O). The pKa value for HCl is -log(Ka) = -log(1.0x10^-14) = 14.

Therefore, the final pH can be calculated as:

pH = 14 + log(373.1/373.1) = 14 + log(1) = 14

This means that the final pH after the electrolysis will remain at 14, which is a basic solution. This is because for every mole of HCl formed, a mole of NaOH is also produced, resulting in no change in the overall concentration of H+ ions and therefore no change in pH.