Расчитать ph 0,1 M раствора CH3COOH k CH3COOH=1,74×10 в минус пятой степени
CH3COOH----> k= 1,74*10⁻⁵ = [H⁺]²/0,1 [H⁺]²=1,74*10⁻⁶ [H⁺]=1,32*10⁻³ pH=2,88
![\frac{[CH3COO^{-}][ H^{+} ]}{0,1 }](https://tex.z-dn.net/?f=+%5Cfrac%7B%5BCH3COO%5E%7B-%7D%5D%5B+H%5E%7B%2B%7D+%5D%7D%7B0%2C1+%7D+ "\frac{[CH3COO^{-}][ H^{+} ]}{0,1 }")