Jawab + cara ya :D
Sebanyak 500 mL larutan NH4OH dengan pH = 11 ( Kb = 10-5 ) diencerkan dengan air sehingga volumenya menjadi 1 L. Hitunglah pH sebelum dan setelah pengenceran !
Sebanyak 500 mL larutan NH4OH dengan pH = 11 ( Kb = 10-5 ) diencerkan dengan air sehingga volumenya menjadi 1 L. Hitunglah pH sebelum dan setelah pengenceran !
Risep
Sebelum pengenceran ya pH = 11
![pH=11 \rightarrow pOH=14-11=3 \rightarrow [OH^-]=10^{-3}~M \\ \\ ~[OH^-]= \sqrt{Kb.M} \\ \\ 10^{-3}= \sqrt{10^{-5}.M} \\ \\ 10^{-6}= {10^{-5}.M \rightarrow M=10^{-1}~M =0,1~M](https://tex.z-dn.net/?f=pH%3D11+%5Crightarrow+pOH%3D14-11%3D3+%5Crightarrow+%5BOH%5E-%5D%3D10%5E%7B-3%7D%7EM+%5C%5C++%5C%5C+%7E%5BOH%5E-%5D%3D+%5Csqrt%7BKb.M%7D+%5C%5C++%5C%5C+10%5E%7B-3%7D%3D+%5Csqrt%7B10%5E%7B-5%7D.M%7D+%5C%5C++%5C%5C+10%5E%7B-6%7D%3D+%7B10%5E%7B-5%7D.M+%5Crightarrow+M%3D10%5E%7B-1%7D%7EM+%3D0%2C1%7EM "pH=11 \rightarrow pOH=14-11=3 \rightarrow [OH^-]=10^{-3}~M \\ \\ ~[OH^-]= \sqrt{Kb.M} \\ \\ 10^{-3}= \sqrt{10^{-5}.M} \\ \\ 10^{-6}= {10^{-5}.M \rightarrow M=10^{-1}~M =0,1~M")
![[OH^-]= \sqrt{Kb.M} \\ \\ = \sqrt{10^{-5}.5.10^{-2}} \\ \\ = \sqrt{5.10^{-7}} \\ \\ =2,24.10^{-3,5} \\ \\ pOH=-log2,24.10^{-3,5}=3,5-log2,24 \\ \\ pH=14-(3,5-log2,24) \\ \\ =10,5+log2,24](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D+%5Csqrt%7BKb.M%7D+%5C%5C++%5C%5C+%3D+%5Csqrt%7B10%5E%7B-5%7D.5.10%5E%7B-2%7D%7D+%5C%5C++%5C%5C+%3D+%5Csqrt%7B5.10%5E%7B-7%7D%7D+%5C%5C++%5C%5C+%3D2%2C24.10%5E%7B-3%2C5%7D+%5C%5C++%5C%5C+pOH%3D-log2%2C24.10%5E%7B-3%2C5%7D%3D3%2C5-log2%2C24+%5C%5C++%5C%5C+pH%3D14-%283%2C5-log2%2C24%29+%5C%5C++%5C%5C+%3D10%2C5%2Blog2%2C24 "[OH^-]= \sqrt{Kb.M} \\ \\ = \sqrt{10^{-5}.5.10^{-2}} \\ \\ = \sqrt{5.10^{-7}} \\ \\ =2,24.10^{-3,5} \\ \\ pOH=-log2,24.10^{-3,5}=3,5-log2,24 \\ \\ pH=14-(3,5-log2,24) \\ \\ =10,5+log2,24")
2 votes
Thanks 2
More Questions From This User See All