larutan 0,2 M XCl (Kb=2.10^-5) akan terhidrolisis sebesar...

Kw = 10^-14
0,2 M = 2×10^-1 M

[H+] = √ (Kw × M)/Kb
[H+] = √ (10^-14 × 2×10^-1)/2×10^-5
[H+] = √ (2×10^-15)/2×10^-5
[H+] = √ 10^-10
[H+] = 10^-5

pH = -log 10^-5
pH = 5 -log 10

[tex]XCl\ 0,2\ M\\XCl\ \textgreater \ X^{+}+Cl^{-}\\ 0,2\ \ \ \ 0,2\ \ \ \ \ \ 0,2\\ \\ karena\ cl\ merupakan\ ion\ sisa\ asam\ kuat\ dan\ x\ ion\ sisa\ basa\ lemah\\ maka\ yang\ akan\ mengalami\ hidrolisis(bereaksi\ dengan\ air)adalah\\ ion\ yang\ lemah.\\ x^{+}+H_{2}O\ \textgreater \ XOH+H^{+}\\ maka\ garam\ ini\ akan\ menjadi\ asam\ karena\ terbentuk\ ion\ H^{+}\\ sehingga\ konsentrasi\ H^{+}\ lebih\ banyak\ dari\ konsentrasi\ OH^{-}.\\ maka\ yang\ dapat\ dicari\ dari\ reaksi\ tersebut\\ \\ H^{+}=\sqrt{kh\times L}[/tex]
[tex]H^{+}=\sqrt{\frac {kw}{kb}\times L}\\ H^{+}=\sqrt{\frac {10^{-14}}{2.10^{-5}}\times 2.10^{-1}}\\ H^{+}=\sqrt{10^{-10}}\\ H^{+}=10^{-5}\\ jika\ ditanya\ derajat\ hidrolisis\ atau\ \alpha\\ \alpha=\sqrt{\frac {kh}{M}}\\ \alpha=\sqrt {\frac {kw}{kb}\times \frac {1}{M}}\\ \alpha=\sqrt {\frac {10^-14}{2.10^{-5}}\times \frac {1}{2.10^{-1}}}\\ \alpha=\sqrt {\frac {10^{-14}}{4.10^{-6}}}\\ \alpha=\sqrt {0,25.10^{-8}}\\ \alpha=\sqrt{\frac {1}{4}.10^{-8}}\\ \alpha=\frac {1}{2}.10^{-4}\\ \alpha=5.10^{-5}[/tex]
[tex]\alpha=0,00005\\ \alpha=0,005\ persen.\\ jika\ ditanya\ ph\\ ph=-log[h^{+}]\\ ph=-log10^{-5}\\ ph=5[/tex]