Diketahui garam KX sebanyak 7,35 gram dilarutkan dalam air higga diperoleh 750 mL larutan dengan pH=9. jika Ka HX=10^-5 maka Ar X adalah...? (Ar H=1,K=39)
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![pH=9 \rightarrow pOH=5 \rightarrow [OH^-]=10^{-5} \\ \\ ~[OH^-]= \sqrt{ \frac{Kw}{Ka}[X^-]} \\ \\ 10^{-5}= \sqrt{ \frac{10^{-14}}{10^{-5}}.[X^-]} \\ \\ 10^{-10}=10^{-9}.[X^-] \\ \\ ~[X^-]=10^{-1}=0,1~M](https://tex.z-dn.net/?f=pH%3D9+%5Crightarrow+pOH%3D5+%5Crightarrow+%5BOH%5E-%5D%3D10%5E%7B-5%7D+%5C%5C++%5C%5C+%7E%5BOH%5E-%5D%3D+%5Csqrt%7B+%5Cfrac%7BKw%7D%7BKa%7D%5BX%5E-%5D%7D+%5C%5C++%5C%5C+10%5E%7B-5%7D%3D+%5Csqrt%7B+%5Cfrac%7B10%5E%7B-14%7D%7D%7B10%5E%7B-5%7D%7D.%5BX%5E-%5D%7D+%5C%5C++%5C%5C+10%5E%7B-10%7D%3D10%5E%7B-9%7D.%5BX%5E-%5D+%5C%5C++%5C%5C+%7E%5BX%5E-%5D%3D10%5E%7B-1%7D%3D0%2C1%7EM "pH=9 \rightarrow pOH=5 \rightarrow [OH^-]=10^{-5} \\ \\ ~[OH^-]= \sqrt{ \frac{Kw}{Ka}[X^-]} \\ \\ 10^{-5}= \sqrt{ \frac{10^{-14}}{10^{-5}}.[X^-]} \\ \\ 10^{-10}=10^{-9}.[X^-] \\ \\ ~[X^-]=10^{-1}=0,1~M")
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