How many grams of $\ce{CH3COONa}$ we need to dissolve in $\pu{2.50 L}$ of water to get a solution with $\mathrm{pH} = 9.00$? Given $K_\mathrm{a}(\ce{CH3COOH}) = 1.75\cdot 10^{-5}$
Like solving a puzzle, I found from the asked $\mathrm{pH}$ the concentration of hydrogen cation and hydroxide anion, and wrote to myself that I need to find the final concentration of $\ce{CH3COONa}$ to seek from it the amount and then the weight.
But my problem was in all the middle steps. I understand the the concentration of hydroxide anions that I found is after the equilibrium, but I miss the point of how to use them to find the concentration on the acetic acid by given the $K_\mathrm{a}$. I'm guessing that I need to get the $K_\mathrm{b}$ from the $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$ equation. But why exactly and how then, I need to find the concentration of $\ce{CH3COONa}$?
Maybe I lack the knowledge and thus the imagination for it.
Support highly appreciated!