Using KOH rather than NaOH will work just fine.

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

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

Please don't hesitate to ask for any clarifications in the comments below, and best of luck!

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.

Please don't hesitate to ask for any clarifications in the comments below, and best of luck!