tentukan ph dari 0,6 gram ch3cooh dilarutkan dalam air 250 ml ka=1.10-5

[tex]mol\ ch3cooh\\ n=\frac {gr}{mr}\\ \\ n=\frac {0,6}{60}\\ n=0,01\ gram/mol\\ molaritas\ =\frac {n}{v}\\ m=\frac {n}{v}\\ m=\frac {0,01}{0,25}\\ m=0,04\ M\\ \\ ch3cooh(asam\ lemah)\\ ch3cooh\ jika\ dilarutkan\ dalam\ air\ maka\ terbentuk\ ion\ ion\ nya,\\ ch3cooh\ \textgreater \ ch3coo^{-}+h^{+}\\ 0,04\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ x \\ \\ H^{+}(x)=\sqrt {Ka\times M}\\ H^{+}=\sqrt {10^{-5}\times 4.10^{-2}}\\ H^{+}=\sqrt {4.10^{-7}}\\ H^{+}=2.10^{-3,5}\\ ph=-log[H^{+}]\\ ph=-log2.10^{-3,5}[/tex]
[tex]ph=-log2-log10^{-3,5}\\ ph=-log10^{-3,5}-log2\\ ph=log10^{3,5}-log2\\ ph=3,5-log2\\ ph=3,5-0,3\\ ph=3,2\\ ph\ \textless \ 7=asam\\ ph\ \textgreater \ 7=basa\\ ph=7\ netral,(air/h2o,garam\ yang\ terbentuk\ dari\ asam\ kuat\ +\\ basa\ kuat)[/tex]