Perbandingan volume dari larutan asam sulfat 0.05M dengan larutan amonia 0.05M yang dapat membentuk larutan penyangga dengan PH=10 adalah...
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![Misalkan\\ V_1=vol\ H_2SO_4\rightarrow\ mol\ H_2SO_4=0,05V_1\\V_2=vol\ NH_3~~~\rightarrow\ mol\ NH_3=0,05V_2\\pH=10\ \rightarrow\ pOH=4\rightarrow\ [OH^-]=10^{-4}\ M\\\\ ~~~~~H_2SO_4~~+~~2NH_3~~~\rightarrow ~~~(NH_4)_2SO_4\\A:0,05V_1~~~~~~~0,05V_2~~~~~~~~~~~~~~-\\R:0,05V_1~~~~~~~~0,1V_1~~~~~~~~~~~~~~~0,05V_1\\S:~~-~~~~~(0,05V_2-0,1V_1)~~~~~~~~~0,05V_1](https://tex.z-dn.net/?f=Misalkan%5C%5C+V_1%3Dvol%5C+H_2SO_4%5Crightarrow%5C+mol%5C+H_2SO_4%3D0%2C05V_1%5C%5CV_2%3Dvol%5C+NH_3%7E%7E%7E%5Crightarrow%5C+mol%5C+NH_3%3D0%2C05V_2%5C%5CpH%3D10%5C+%5Crightarrow%5C+pOH%3D4%5Crightarrow%5C+%5BOH%5E-%5D%3D10%5E%7B-4%7D%5C+M%5C%5C%5C%5C+%7E%7E%7E%7E%7EH_2SO_4%7E%7E%2B%7E%7E2NH_3%7E%7E%7E%5Crightarrow+%7E%7E%7E%28NH_4%29_2SO_4%5C%5CA%3A0%2C05V_1%7E%7E%7E%7E%7E%7E%7E0%2C05V_2%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E-%5C%5CR%3A0%2C05V_1%7E%7E%7E%7E%7E%7E%7E%7E0%2C1V_1%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E0%2C05V_1%5C%5CS%3A%7E%7E-%7E%7E%7E%7E%7E%280%2C05V_2-0%2C1V_1%29%7E%7E%7E%7E%7E%7E%7E%7E%7E0%2C05V_1 "Misalkan\\ V_1=vol\ H_2SO_4\rightarrow\ mol\ H_2SO_4=0,05V_1\\V_2=vol\ NH_3~~~\rightarrow\ mol\ NH_3=0,05V_2\\pH=10\ \rightarrow\ pOH=4\rightarrow\ [OH^-]=10^{-4}\ M\\\\ ~~~~~H_2SO_4~~+~~2NH_3~~~\rightarrow ~~~(NH_4)_2SO_4\\A:0,05V_1~~~~~~~0,05V_2~~~~~~~~~~~~~~-\\R:0,05V_1~~~~~~~~0,1V_1~~~~~~~~~~~~~~~0,05V_1\\S:~~-~~~~~(0,05V_2-0,1V_1)~~~~~~~~~0,05V_1")
![$ \begin {align} [H^+]&=K_b\times\frac{[NH_3]}{[NH_4^+]}\\\\10^{-4}&=10^{-5}\times\frac{0,05V_2-0,1V_1}{0,1V_1}\\\\V_1&=0,05V_2-0,1V_1\\1,1V_1&=0,05V_2\\\\\frac{V_1}{V_2}&=\frac{0,05}{1,1}\\&=\frac{5}{110}\\&=\boxed{\boxed{\bold{\frac{1}{22}}}} \end](https://tex.z-dn.net/?f=%24+%5Cbegin+%7Balign%7D+%5BH%5E%2B%5D%26%3DK_b%5Ctimes%5Cfrac%7B%5BNH_3%5D%7D%7B%5BNH_4%5E%2B%5D%7D%5C%5C%5C%5C10%5E%7B-4%7D%26%3D10%5E%7B-5%7D%5Ctimes%5Cfrac%7B0%2C05V_2-0%2C1V_1%7D%7B0%2C1V_1%7D%5C%5C%5C%5CV_1%26%3D0%2C05V_2-0%2C1V_1%5C%5C1%2C1V_1%26%3D0%2C05V_2%5C%5C%5C%5C%5Cfrac%7BV_1%7D%7BV_2%7D%26%3D%5Cfrac%7B0%2C05%7D%7B1%2C1%7D%5C%5C%26%3D%5Cfrac%7B5%7D%7B110%7D%5C%5C%26%3D%5Cboxed%7B%5Cboxed%7B%5Cbold%7B%5Cfrac%7B1%7D%7B22%7D%7D%7D%7D+%5Cend+ "$ \begin {align} [H^+]&=K_b\times\frac{[NH_3]}{[NH_4^+]}\\\\10^{-4}&=10^{-5}\times\frac{0,05V_2-0,1V_1}{0,1V_1}\\\\V_1&=0,05V_2-0,1V_1\\1,1V_1&=0,05V_2\\\\\frac{V_1}{V_2}&=\frac{0,05}{1,1}\\&=\frac{5}{110}\\&=\boxed{\boxed{\bold{\frac{1}{22}}}} \end")
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