Similar to acetic acid, benzoic acid (C6H5COOH) is a weak acid thatis commonly u

Question

Similar to acetic acid, benzoic acid (C6H5COOH) is a weak acid thatis commonly used as an antimicrobial agemt and found in toothpasteand mouthwashes, cosmetics and deodrants. Buffered benzoic acid isused in both food and soft drinks to imporve their flavor. Ka forbenzoic acid is 6.5 x 10-5.
Calculate the pH for the mixture that contains 10.00ml of0.8000M benzoic acid (C6H5COOH) AND 40.00ML OF 0.2000M NaOH.
Please show how to check assumption that is made duringcalculation if assumption is made.
Step by step procedure to attaining an answer abovewould be truly appreciated. Trying to study for a midterm tomorrow!Thank you!
Calculate the pH for the mixture that contains 10.00ml of0.8000M benzoic acid (C6H5COOH) AND 40.00ML OF 0.2000M NaOH.
Please show how to check assumption that is made duringcalculation if assumption is made.
Step by step procedure to attaining an answer abovewould be truly appreciated. Trying to study for a midterm tomorrow!Thank you!

Explanation / Answer

Given Ka for benzoic acid is 6.5 x 10-5 No. of moles of benzoic acid n = molarity * volume                                                 =0.8 * 10                                                = 8 mol No. of moles of NaOH, n' = molarity * volume                                           = 0.2 * 40                                           = 8 mol                                       C6H5COOH     +   NaOH  ----------->     C6H5COONa + H2O initialmoles                               8                      8                                      0
Since  C6H5COOH is weak acid & both acid andbase are of same conc. then ([salt] / [ acid] ) =1
                     pH = pKa + log ( [salt] / [ acid])                            = - log Ka + log 1                            = 4.187 No. of moles of benzoic acid n = molarity * volume                                                 =0.8 * 10                                                = 8 mol No. of moles of NaOH, n' = molarity * volume                                           = 0.2 * 40                                           = 8 mol                                       C6H5COOH     +   NaOH  ----------->     C6H5COONa + H2O initialmoles                               8                      8                                      0
Since  C6H5COOH is weak acid & both acid andbase are of same conc. then ([salt] / [ acid] ) =1
                     pH = pKa + log ( [salt] / [ acid])                            = - log Ka + log 1                            = 4.187 initialmoles                               8                      8                                      0
Since  C6H5COOH is weak acid & both acid andbase are of same conc. then ([salt] / [ acid] ) =1
                     pH = pKa + log ( [salt] / [ acid])                            = - log Ka + log 1                            = 4.187

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