Jelaskan perbandingan volume dari larutan asam sulfat 0,05 M dengan larutan amonia 0,05 M yang dapat membentuk larutan penyangga dengan pH = 10 adalah
Tolong ykk
Tolong ykk
fatihahalhusna
![~~~~~~2NH_3~~+~~~H_2SO_4\rightarrow (NH_4)_2SO_4\\A:~~0,05V_1~~~~~~~0,05V_2~~~~~~~~~-\\R:~~~0,1V_2~~~~~~~0,05V_2~~~~~~~0,05V_2\\S:(0,05V_1-0,1V_2)-~~~~~~~~~0,05V_2\\\\pH=10\rightarrow pOH=4\rightarrow [OH^-]=10^{-5}\ M\\\\{[OH^-]}=K_b\frac{mol\ NH_3}{mol\ NH_4^+}\\\\10^{-5}=10^{-5}\times\frac{0,05V_1-0,1V_2}{0,1V_2}\\\\0,05V_1-0,1V_2=0,1V_2\\\\\displaystyle \frac{V_2}{V_1}=\frac{0,05}{0,2}=\boxed{\boxed{\bold{\frac{1}{4}}}}](https://tex.z-dn.net/?f=%7E%7E%7E%7E%7E%7E2NH_3%7E%7E%2B%7E%7E%7EH_2SO_4%5Crightarrow+%28NH_4%29_2SO_4%5C%5CA%3A%7E%7E0%2C05V_1%7E%7E%7E%7E%7E%7E%7E0%2C05V_2%7E%7E%7E%7E%7E%7E%7E%7E%7E-%5C%5CR%3A%7E%7E%7E0%2C1V_2%7E%7E%7E%7E%7E%7E%7E0%2C05V_2%7E%7E%7E%7E%7E%7E%7E0%2C05V_2%5C%5CS%3A%280%2C05V_1-0%2C1V_2%29-%7E%7E%7E%7E%7E%7E%7E%7E%7E0%2C05V_2%5C%5C%5C%5CpH%3D10%5Crightarrow+pOH%3D4%5Crightarrow+%5BOH%5E-%5D%3D10%5E%7B-5%7D%5C+M%5C%5C%5C%5C%7B%5BOH%5E-%5D%7D%3DK_b%5Cfrac%7Bmol%5C+NH_3%7D%7Bmol%5C+NH_4%5E%2B%7D%5C%5C%5C%5C10%5E%7B-5%7D%3D10%5E%7B-5%7D%5Ctimes%5Cfrac%7B0%2C05V_1-0%2C1V_2%7D%7B0%2C1V_2%7D%5C%5C%5C%5C0%2C05V_1-0%2C1V_2%3D0%2C1V_2%5C%5C%5C%5C%5Cdisplaystyle+%5Cfrac%7BV_2%7D%7BV_1%7D%3D%5Cfrac%7B0%2C05%7D%7B0%2C2%7D%3D%5Cboxed%7B%5Cboxed%7B%5Cbold%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%7D "~~~~~~2NH_3~~+~~~H_2SO_4\rightarrow (NH_4)_2SO_4\\A:~~0,05V_1~~~~~~~0,05V_2~~~~~~~~~-\\R:~~~0,1V_2~~~~~~~0,05V_2~~~~~~~0,05V_2\\S:(0,05V_1-0,1V_2)-~~~~~~~~~0,05V_2\\\\pH=10\rightarrow pOH=4\rightarrow [OH^-]=10^{-5}\ M\\\\{[OH^-]}=K_b\frac{mol\ NH_3}{mol\ NH_4^+}\\\\10^{-5}=10^{-5}\times\frac{0,05V_1-0,1V_2}{0,1V_2}\\\\0,05V_1-0,1V_2=0,1V_2\\\\\displaystyle \frac{V_2}{V_1}=\frac{0,05}{0,2}=\boxed{\boxed{\bold{\frac{1}{4}}}}")
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fatihahalhusna
Nilai Kb tidak disebutkan disoal saya buat nilai Kb=10^-5