Chemistry, 14.07.2019 16:30, thomasmurphy200
The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a solution of nac6h5coo if the ph is 9.04?
Answers: 1
![[OH^{-}] = 10^{^{-pOH}}](/tpl/images/0089/2713/549cf.png)
![[OH^{-}] = 10^{^{-4.96}}](/tpl/images/0089/2713/ff5c3.png)
![[OH^{-}] = 1.1\times 10^{-5}](/tpl/images/0089/2713/97cf5.png)
![kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}](/tpl/images/0089/2713/6f928.png)
and value of C6H5COOH is equal to OH-![1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}](/tpl/images/0089/2713/af234.png)
![[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}](/tpl/images/0089/2713/c4847.png)
![[C6H5COO^{-}] = 0.76 M](/tpl/images/0089/2713/e77f3.png)
or 
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The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are p...
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