Open Book (75 points) Some might be more difficult than others. Read over all an

Question

Open Book (75 points) Some might be more difficult than others. Read over all and start with the one onapproach weigh You must show your work step-by-step to get corresponding credit. Your lo more than the final numeric results. gic and approach weigh 1. A 500.00 mL solution is prepared by dissolvi 6.20 sodium acetate in water. The pKa of the conjugate acid (acetic acid) is 4.75. Calculate the solution pH and the concentration of acetate and acetic tion is prepared by dissolving 6.20 sodium acetate in water. The pKa of the dissolvig acid (in molarity mol/L). Use the molecular weight of sodium acetate of 82.0 g/mol. 20 points

Explanation / Answer

Q1

expect hydrolysis

CH3COO-(aq) + H2O(l) <-> CH3COOH + OH-(aq)

Let HA --> CH3COOH and A- = CH3COO - 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-]

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

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

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

M = mol/V = (6.2/82)/0.5 = 0.1512

5.55*10^-10 = x*x/(0.1512-x)

solve for x

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

solve for x with quadratic equation

x = OH- =9.16*10^-6

[OH-]  =9.16*10^-6

pOH = -log(OH-) = -log(9.16*10^-6= 5.04

pH = 14-5.04= 8.96

pH = 8.96

[OH-] = 0 + x = 9.16*10^-6

[HA] = 0 + x = 9.16*10^-6 (acetic ciad)

[A-] = M - x = (6.2/82/0.5) - 9.16*10^-6 = 0.1512 the acetate

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