Chemistry, 26.07.2019 22:00 beiachapman122
What is the ph of a 0.40 m solution of benzoic acid, c6h5cooh? (the ka value for benzoic acid is 6.3×10−5?
Answers: 1
![Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}](/tpl/images/0136/4833/06942.png)
![Ka= \frac{x*x}{[C_6H_5COOH]_i-x} \\ \\ x= \sqrt{[C_6H_5COOH]_i*Ka}= \sqrt{0,40M*6,3*10^{-5} } \\ \\ x=0,00502M=[H_3O^{+} ]](/tpl/images/0136/4833/de548.png)
![pH=-log[H_3O^{+}] =-log (0,00502M)=2,30](/tpl/images/0136/4833/e0116.png)
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What is the ph of a 0.40 m solution of benzoic acid, c6h5cooh? (the ka value for benzoic acid is 6....
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