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Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 M HC7H5O2 with 100.0 mL of 0.30 M NaC7H5O2. The Ka for HC7H5O2 is 6.5 × 10-5.
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we know that pH=pKa + log(base/acid) and here we know that pKa=-log(Ka) and we know that base content and the acid content. its simple from here we just plug these values in. can you do this and tell me what you get?
The new concnetrations are just moles over total volume.\[{.10M \times .150L \over .150L + .100L }= .06MHC _{7} H _{5} O _{2}\]\[pH=-\log(6.5x10^{-5})+\log{[.06]\over[.12]}\]
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