1. a. How many milliliters of O.0998 M NaOH are required to \"neutralize\" (reac
ID: 588524 • Letter: 1
Question
1. a. How many milliliters of O.0998 M NaOH are required to "neutralize" (react completely) with 0.3052 g of benzoic acid, C6HsCOOH, (Molar mass = 122.12g/mol). b. What is the pH of the resulting solution after the "neutralization" reaction (that is after the NaOH solution has been added to the benzoic acid)? pKa of benzoic acid is 4.20. (You may assume the solid benzoic acid sample was dissolved in 25.0 mL of distilled, deionized water before being reacted with the base to assist in determining the volume of the solution after the reaction is complete.)Explanation / Answer
1)
a) moles of benzoic acid = 0.3052 / 122.12 = 0.0025
0.0025 moles NaOH must be added to neutralize.
V = n /M = 0.0025 / 0.0998
V = 0.025 L
V = 25.0 mL
volume of NaOH added = 25.0 mL
b) total acid becomes aalt
[acid] = 0.0025 / 0.0250.1 M
after reacts with NaOH
[salt] = 0.0025 / 0.05
[salt] = 0.05 M
pH = 1/2[pKw + pka + log C]
pH = 1/2 [14 + 4.20 + log 0.05]
pH = 8.45
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