Answer:
None of the species in the equation have undergone either oxidation or reduction
Explanation:
The easiest way to see if oxidation or reduction has happened is to compare oxidation numbers of each and every species before and after the reaction.
Calcium is two before the reaction and two after the reaction
Hydrogen is -1 before the reaction and -1 after the reaction
Sodium is one before the reaction and one after the reaction
Iodine is -1 before the reaction and -1 after the reaction.
For an oxidation to happen an increase in oxidation number has to happen.
For a reduction to happen, a decrease in oxidation number has to happen. None have happened
N[Fe(NO₃)₃]=15.0 mol
n(O)=3×3×n[Fe(NO₃)₃]=9n[Fe(NO₃)₃]
n(O)=9×15.0=135.0 mol
First, we have to see how K2O behaves when it is dissolved in water:
K2O + H20 = 2 KOH
According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:
c(KOH) : c(K2O) = 1 : 2
So,
c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)
Now we can calculate pH:
pOH= -log c(OH⁻) = -log 0.01 = 2
pH= 14-2 = 12
Answer:
Explanation:
a).
conc of Ca²⁺ =0.0025 M
pCa = -log(0.0025) = 2.6
logK,= 10.65 So lc = 4.47 x 10.
Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is
=0.81
So the Conditional Formation constant=
=0.81x 4.47 x10¹⁰
=3.62x10¹⁰
b)
At Equivalence point:
Ca²⁺ forms 1:1 complex with EDTA At equivalence point,
Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol
Number of moles of EDTA= 0.125 mol
Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL
V e= 25.00 mL
At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.
![[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M](https://tex.z-dn.net/?f=%5BCaY%5E%7B2-%7D%5D%20%3D%20%5Cfrac%7BInitial%2Cmoles%2Cof%2C%20Ca%5E%7B2%2B%7D%7D%7BTotal%2CVolume%7D%20%3D%20%5Cfrac%7B0.125mol%7D%7B%2850.00%2B25.00%29mL%7D%20%3D%200.001667M)

Ca²⁺ + Y⁴ ⇄ CaY²⁻
Initial 0 0 0.001667
change +x +x -x
equilibrium x x 0.001667 - x
![{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\](https://tex.z-dn.net/?f=%7BK%5E%27%7D_f%20%3D%20%5Cfrac%7B%5BCaY%5E%7B2-%7D%5D%7D%7B%5BCa%5E%7B2%2B%7D%5D%5BY%5E4%5D%7D%3D%5Cfrac%7B0.001667-x%7D%7Bx.x%7D%20%3D%5Cfrac%7B0.001667-x%7D%7Bx%5E2%7D%5C%5C%5C%5Cx%5E2%20%3D%20%5Cfrac%7B0.001667-x%7D%7B%7BK%5E%27%7D_f%7D%5C%5C%20%5C%5C)

x = 2.15×10⁻⁷
[Ca+2] = 2.15x10⁻⁷ M
pca = —log(2 15x101= 6.7