8. The acetate ion, CH,COO, reacts with water as a weak base: CH3COO-(aq) + H20(

Question

8. The acetate ion, CH,COO, reacts with water as a weak base: CH3COO-(aq) + H20(c) CH3COOH(aq) + OH-(aq), with K-5.7 x 10-10 at 25°C. If sodium acetate were dissolved in water to make a 0.65 M solution, what would be the resulting concentration of OH? 9. For the phase transition H2O(C) H20(g) at 298 K,AH_ 44.0 kJ/mol and So-119 J/mol x K. What is the value of AG at 298 K when the partial pressure of water is 42.8 torr? 10. Given the following two measurements of the equilibrium constant for a reaction, calculate //o for the reaction. T, °C 10.0 20.0 2.1 x 103 8.1 x 104

Explanation / Answer

8)

Lets write the dissociation equation of CH3COO-

CH3COO- +H2O -----> CH3COOH + OH-

0.65 0 0

0.65-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.7*10^-10)*0.65) = 1.925*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.925*10^-5 M

So, [OH-] = x = 1.925*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.925*10^-5)

= 4.72

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.72

= 9.28

Answer: 9.28

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