Chemistry, 11.07.2019 12:20 Epicgible8136
A0.070 m solution of the salt nab has a ph of 9.60. calculate the ph of a 0.010 m solution of hb. or, what is the kb of hb?
Answers: 1
![pH=-log([H^+])\\\\ H^+=10^{-pH}=10^{-9.60}=2.51x10^{-10}\\OH^-=\frac{Kw}{[H^+]} =\frac{1x10^{-14}}{2.51x10^{-10}} =3.98x10^{-5}](/tpl/images/0077/0554/65ff2.png)
![[OH^-]=[HB]=3.98x10^{-5}M](/tpl/images/0077/0554/b3ba7.png)
![Kb=\frac{[HB][OH^-]}{[B^-]}=\frac{(3.98x10^{-5})^2}{0.070-3.98x10^{-5}} =2.264x10^{-8}](/tpl/images/0077/0554/4b31b.png)
![Ka=\frac{[H^+][B^-]}{[HB]}\\4.42x10^{-6}=\frac{x^2}{0.010-x} \\x=2.08x10^-4M](/tpl/images/0077/0554/0af12.png)
![[H^+]=2.08x10^{-4}\\pH=-log(2.08x10^{-4})\\pH=3.68](/tpl/images/0077/0554/e8d31.png)
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A0.070 m solution of the salt nab has a ph of 9.60. calculate the ph of a 0.010 m solution of hb. o...
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