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11 Dec 2019
Ka of benzoic acid ( HC7H5O2) = 6.5 x 10^-5 . 0.288 g of sodium benzoate (NaC7H5O2 , Mw = 144) is added 10.0 mL of a 0.0500 M HCl solution. Find the pH of the resultant solution.
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Ka of benzoic acid ( HC7H5O2) = 6.5 x 10^-5 . 0.288 g of sodium benzoate (NaC7H5O2 , Mw = 144) is added 10.0 mL of a 0.0500 M HCl solution. Find the pH of the resultant solution.
SHOW ALL WORK AND EXPLAIN!
Thank you!
Tod ThielLv2
13 Dec 2019