Sebanyak 0,1 mol 
( 
= 
) dicampurkan dengan 0,05 mol 
hingga volume 1 liter. pH larutan yang dihasilkan adalah...
A. 5
B. 5 + log 2
C. 5 – log 2
D. 9 + log 2
E. 9 – log 2
A. 5
B. 5 + log 2
C. 5 – log 2
D. 9 + log 2
E. 9 – log 2
Risep
Larutan Penyangga
![[OH^-]=Kb. \frac{[NH_4OH]}{[NH_4Cl]} \\ \\ =10^{-5}. \frac{0,1/1}{0,05/1} \\ \\ =10^{-5}.2 \\ \\ =2.10^{-5} \\ \\ \\ pOH=-log 2.10^{-5} \\ =5-log2 \\ \\ pH=14-pOH \\ =14-(5-log2) \\ =9+log2](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3DKb.+%5Cfrac%7B%5BNH_4OH%5D%7D%7B%5BNH_4Cl%5D%7D+%5C%5C++%5C%5C+%3D10%5E%7B-5%7D.+%5Cfrac%7B0%2C1%2F1%7D%7B0%2C05%2F1%7D+%5C%5C++%5C%5C+%3D10%5E%7B-5%7D.2+%5C%5C++%5C%5C+%3D2.10%5E%7B-5%7D+%5C%5C++%5C%5C++%5C%5C+pOH%3D-log+2.10%5E%7B-5%7D+%5C%5C+%3D5-log2+%5C%5C++%5C%5C+pH%3D14-pOH+%5C%5C+%3D14-%285-log2%29+%5C%5C+%3D9%2Blog2 "[OH^-]=Kb. \frac{[NH_4OH]}{[NH_4Cl]} \\ \\ =10^{-5}. \frac{0,1/1}{0,05/1} \\ \\ =10^{-5}.2 \\ \\ =2.10^{-5} \\ \\ \\ pOH=-log 2.10^{-5} \\ =5-log2 \\ \\ pH=14-pOH \\ =14-(5-log2) \\ =9+log2")
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