You are asked to prepare a ph=4.00 buffer starting from 1.50 l of 0.0200 m solution of benzoic acid (c6h5cooh) and an excess of sodium benzoate (c6h5coona). you may want to reference (pages 721 - 728) section 17.2 while completing this problem. part a what is the ph of the benzoic acid solution prior to adding sodium benzoate?

pH = 2.96

Explanation:

∴ Ka = 6.5 E-5 = [H3O+][C6H5COO-] / [C6H5COOH]

prior to adding C6H5COONa:

⇒ C C6H5COONa = 0 M = [Na+]

∴ mass balance:

⇒ C C6H5COOH = [C6H5COOH] + [C6H5COO-] = 0.0200 M

⇒ [C6H5COOH] = 0.0200 - [C6H5COO-]

charge balance:

⇒ [H3O+] = [C6H5COO-]

⇒ [C6H5COOH] = 0.0200 - [H3O+]

⇒ Ka = [H3O+]² / (0.0200 - [H3O+]) = 6.5 E-5

⇒ [H3O+]² + 6.5 E-5[H3O+] - 1.3 E-6 = 0

⇒ [H3O+] = 1.108 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.9554 = 2.96

if you mean hydrogen and oxygen then water is the product

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.