Please help!! I\'m lost with these questions!!! Please explain the steps! 8.6 lo
ID: 508601 • Letter: P
Question
Please help!! I'm lost with these questions!!!Please explain the steps! 8.6 low) .Me is some metal. Calculate the pH Value of the saturerted soluuton of MeroH)e m Pure water e)The solubility of MeloH, m puce water is: bout the same e) Cant Hell less than than the solubihty of MetOH. in HI solutio o. C The value of Ka for some acid CHA at 25 C is Ka 4.5 a By using the value of Ka calculate AGO for Hhe dissociation of HA b bat the value ot AG at equilibrium? CS whoct ot AG a+ uilibrium when CH-1 to uo CHAI a. 4 uo 3M, and CAI 40 AM
Explanation / Answer
A) Me(OH)2 (s)< --------> Me2+ (aq) + 2OH-(aq)
Ksp = [Me2+][OH-]^2
At saturated condition
Qsp = Ksp
Qsp = reaction quotient (actual ionic product)
Therefore, Qsp = 8.6×10^-14
8.6 × 10^-14 = [Me2+][OH-]^2
If we put solubility of Me(OH)2 as S then
8.6 × 10^-14= 1S × (2S)^2
4S^3 = 8.6×10^-14
S^3 = 2.15 × 10^-14
S = 2.78×10^-5
[OH-] = 2S
[ OH - ] = 2 × 2.78 × 10^-5
= 5.56×10^-5M
pOH = -log[OH-]
= - log(5.56×10^-5)
= 4.25
pH + pOH = 14
pH = 14 - pOH
= 14 - 4.25
= 9.75
B) Me(OH)2(s) < --------> Me2+(aq) + 2OH-(aq)
HI is an acid and it will give H+ ions
The H+ ions react with OH- ions of the above equilibrium , so OH- ions removed from the equilibrium
According to Le chatlier principle , the equilibrium will shift to the right and give OH- ions to compensate the reduced OH- ions due to reaction with H+ ions.So , the solubility of Me(OH)2 is increased.
Thus, the solubility of Me(OH)2 in pure water is less than the solubility of Me(OH)2 in HI solution.
C) HA <---------> H+ + A-
Ka = [H+][A-]/[HA] = 4.5× 10^-4
a) G° = - RTlnK
Where, R is gas constant = 8.314( J/mol K )
K is equilibrium constant , Ka =4.5×10^-4
T = Kelvin temperature, 25°C = 273.15 +25 = 298.15K
Substituting the values
G° = - 8.314 ( J/mol K) × 298.15K ×ln (4.5×10^-4)
= -8.314 ( J/mol K) × 298.15K × 2.303 × log (4.5 × 10^-4)
= - 5708.72 (J/mol) × (-3.347)
= 19107J/mol
= 19.11 KJ/mol
b) G = G° + RTlnRTlnQ
At equilibrium Q = K
G = -RTlnK + RTlnK =0
c) Q= [H+][A-]/[HA]
Q is reaction quotient
Q= ((0.1×10^-2M)×(4.0×10^-4M))/2.4×10^-3M
= 1.7 × 10^-4
G = G° + RTlnQ
= 19107J/mol + 8.314 ×298.15 ×2.303 log(1.7×10^-4)
= 19107J/mol - 21519J/mol
= -2412 J/mol
= -2.41KJ/mol