Sebanyak 10,7 gram NH4Cl dilarutkan dalam air hingga bervolume 0,5 L.pH larutan terseut adalah .............Kb NH3=10^-5
fatihahalhusna
![Mr\ NH_4Cl=53,5\\V=0,5\ L=500\ mL\\\\M=\frac{10,7}{53,5}\times\frac{1000}{500}=0,4\ M\\\\{[H^+]}=\sqrt{\displaystyle \frac{K_w\times M}{K_b}}=\displaystyle \sqrt{\frac{10^{-14}\times 0,4}{10^{-5}}}=2\times10^{-5}\ M\\\\\boxed{\boxed{\bold{pH=5-\log2}}}](https://tex.z-dn.net/?f=Mr%5C+NH_4Cl%3D53%2C5%5C%5CV%3D0%2C5%5C+L%3D500%5C+mL%5C%5C%5C%5CM%3D%5Cfrac%7B10%2C7%7D%7B53%2C5%7D%5Ctimes%5Cfrac%7B1000%7D%7B500%7D%3D0%2C4%5C+M%5C%5C%5C%5C%7B%5BH%5E%2B%5D%7D%3D%5Csqrt%7B%5Cdisplaystyle+%5Cfrac%7BK_w%5Ctimes+M%7D%7BK_b%7D%7D%3D%5Cdisplaystyle+%5Csqrt%7B%5Cfrac%7B10%5E%7B-14%7D%5Ctimes+0%2C4%7D%7B10%5E%7B-5%7D%7D%7D%3D2%5Ctimes10%5E%7B-5%7D%5C+M%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7B%5Cbold%7BpH%3D5-%5Clog2%7D%7D%7D "Mr\ NH_4Cl=53,5\\V=0,5\ L=500\ mL\\\\M=\frac{10,7}{53,5}\times\frac{1000}{500}=0,4\ M\\\\{[H^+]}=\sqrt{\displaystyle \frac{K_w\times M}{K_b}}=\displaystyle \sqrt{\frac{10^{-14}\times 0,4}{10^{-5}}}=2\times10^{-5}\ M\\\\\boxed{\boxed{\bold{pH=5-\log2}}}")
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