Let’s start with our generic acid equilibrium constant equation;
Ka=[H+][A−][HA]Ka=[H+][A−][HA]
Which in our case is
Ka=[H+][CH3COOH−][CH3COOH]Ka=[H+][CH3COOH−][CH3COOH]
We can solve this using an Ice Table
Plugging back into our original equation yields
1.8*10^-^5=\frac{[x][x]}{[.1-x]}1.8*10^-^5=\frac{[x][x]}{[.1-x]}
Solving gives us an X value of 0.00133267. (Just plug it into Wolfram Alpha). Since this value is equal to our [H+][H+] we can take the -log of it, giving a pH of 2.875
