PH value of saturated solution of some base Me(OH): is pH 10.56. at 25 degree C.

Question

PH value of saturated solution of some base Me(OH): is pH 10.56. at 25 degree C. The equilibrium constant expression for the dissolving of LaF_3, in water is: K_sp=[Me^2+] [OH"]^2 a) Write the balanced equation for the dissolving of solid Me(OH)_2(s) in water. Show state of all particles (1, s. g .aq). b) Calculate the value of Ksp for Me(OH)_2 At 25degree C The value of Ka for some acid (HA) is Ka=1.8 10^-5 a) By using the value of Ka calculate AG^0 for the dissociation of the acid b) What is the value of delta G at equilibrium? c) What is the value of delta G at equilibrium when [H^+]=6.0 10^-2M, [HA]=0.20M and [A]=6.0 10^-4M?

Explanation / Answer

Question 1.

a)

The equation we require is:

Me(OH)2(s) <--> Me+2(aq) + 2OH-(aq)

b)

for Ksp, we need concentration

pH = 10.56, pOH = 14-10.56= 3.44

[OH-] = 10^-poh = 10^-  3.44 = 0.000363

note that

[M+2] = 1/2*[OH-] due to sotichiometry

so

Ksp = [M+2][OH-]^2

Ksp = 1/2*[OH-] * [OH-]^2

Ksp = 1/2*[OH-]^3

Ksp = 1/2*(0.000363^3)

Ksp = 2.391*10^-11

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