The ph of a 0.1 m ch3cooh solutions is

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

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