pH is equivalent to the negative log of the concentration of
H in molarity, that is:
pH = - log [H]
so the concentration is:
8.05 = - log [H]
[H] = 8.91 x 10^-9 M
the absolute error is:
8.07 = - log [H’]
[H’] = 8.51 x 10^-9 M
absolute error = 8.91 x 10^-9 M - 8.51 x 10^-9 M = 0.4 x
10^-9 M
Therefore:
<span>8.91 x 10^-9 M ± 0.4 x 10^-9 M</span>
Delta Go = -RTlnKeq
delta Go = 5.95 kJ/mole = 5.95 X 1000 = 5950 J/mole ( 1 kj = 1000 J )
putting the values and finding Keq
5950 = -8.314 X 298 X ln Keq
ln Keq = -5950 / 2477.572 = -2.4015
Keq = e^-2.402 = 0.0905
suppose the equilibrium reaction is :-
chair 1 <--------> chair 2
now as Keq is less than 1 ....so chair 1 will be more stable
Keq = [chair2]/[chair 1 ] = 0.0905
this means that [chair 2] ~ 0.0905 and [chair 1] ~ 1
[total] = [chair 2] + [chair 1] ~ 1 + 0.0905= 1.0905
percentage of chair 1 = [chair 1] / [total] = 1 / 1.0905 X 100 = 91.70 %
Answer:
a) [ Ca2+ ] = 3.347 E-4 mol/L
b) [ Ca2+ ] = 1.5 E-8 mol/L
Explanation:
S S 2S......in the equilibrium
⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S )² = 4S³
⇒ S = ∛ ( 1.5 E-10 / 4 )
⇒ S = ∛ 3.75 E-11
⇒ S = 3.347 E-4 mol/L
⇒ [ Ca2+ ] = S = 3.347 E-4 mol/L
b) NaF ↔ Na+ + F-
0.10 M 0.10 0.10
S S 2S + 0.10
⇒ Ksp = 1.5 E-10 = [ Ca2+ ] * [ F- ]² = S * ( 2S + 0.10 )²
∴∴ the Concentration: 0.10 M >>>> Ksp ( 1.5 E-10 ), son we can despise S as adding.
⇒ 1.5 E-10 = S * ( 0.10 )² = 0.01 S
⇒ S = 1.5 E-10 / 0-01
⇒ S = 1.5 E-8 mol/L
⇒ [ Ca2+ ] = S = 1.5 E-8 mol/L
Answer:
The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.
Explanation:
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.
POH indicates the concentration of hydroxyl ions [OH-] present in a solution and is defined as the negative logarithm of the activity of hydroxide ions (that is, the concentration of OH- ions):
pOH= -log [OH-]
A solution has a pOH of 4.90. Replacing in the definition of pOH:
4.90= -log [OH-]
Solving:
-4.90= log [OH-]
1.26*10⁻⁵ M= [OH-]
<u><em>The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.</em></u>
