How many milliliters of 0.0998 M NaOH are required to \"neutralize\" (react comp

Question

How many milliliters of 0.0998 M NaOH are required to "neutralize" (react completely) with 0.3052 g of benzoic acid, . a. C6HsCOOH, (Molar mass = 122.12g/mol). moles Bene Aco 30S 122 113/ 1 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

expect hydrolysis

Let HA --> benzoic acid and A- = benzoate ion for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(10^-4.20) = 1.584*10^-10

Kb = [HA ][OH-]/[A-]

get M in equilbirium

Vtotal = Vacid + Vbase = 25+ 25 = 50 mL

[A-] = mmol/V = 0.0025/50 = 0.05

1.584*10^-10 = x*x/(0.05-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =2.814*10^-6

[OH-]  =2.814*10^-6

pOH = -log(OH-) = -log(2.814*10^-6) = 5.55

pH = 14-5.55= 8.45

pH = 8.45

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