# Will KOH work in place of NaOH as the electolyte in the electrolysis of water?

I am making a hydrogen and oxygen generator by electrolysis water. I found out that using salt as an electrolyte creates a compound of chlorine and oxygen with my electrodes clouding up the water and using all of the oxygen from the reaction. Bubbles are coming off of only one electrode. In my research I found that sodium hydroxide is a better electrolyte because it doesn't create chlorine gas and shouldn't react with my electrodes. Unfortunately I don't have any, but I am making lye ($\ce{KOH}$) from some hardwood ashes, will this work in place of $\ce{NaOH}$? I'm not sure if the potassium ions will conduct electricity in the same way. Thanks!

• Yes, they should work the same. Jun 7 '17 at 18:10
• There is a difference in conductivity between NaOH and KOH due to $\ce{Na+}$ having more spheres of hydration. I would look into conductivity values for the concentration solution you plan to use and see what the difference in conductivity is and determine if it will be an issue. Jun 7 '17 at 18:56
• You will get a number of salts when you extract your ashes. Fractional crystallisation could be used to isolate them if it becomes a problem but you need to accept that it is not just one salt in wood fire ash. Feb 1 '18 at 9:02

Using $\ce{KOH}$ rather than $\ce{NaOH}$ will work just fine.

Solutions of $\ce{KOH}$ are actually about $\pu{5\%}$ more conductive than solutions of $\ce{NaOH}$ at the same molar concentration. In terms of mass concentration, a given mass of $\ce{KOH}$ will have about $\pu{30\%}$ fewer equivalents of reactant than the same mass of $\ce{NaOH}$ because of the difference in molecular weights.

The combination of the difference in molar conductivity and molecular weights between $\ce{NaOH}$ and $\ce{KOH}$ just means that a given concentration of $\ce{KOH}$ should be about $\pu{35\%}$ less conductive than the same mass of $\ce{NaOH}$. Of course you can increase the concentration of $\ce{KOH}$ relative to the amount of $\ce{NaOH}$ you had planned to use if you like.