Answer:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Explanation:
Chemical equation:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Balance chemical equation:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Ionic equation
Ag⁺(aq)+ NO₃⁻(aq) + Na⁺(aq)+ Cl⁻(aq) → AgCl(s) + Na⁺(aq)+ NO₃⁻(aq)
Net ionic equation:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
The NO₃⁻((aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The AgCl can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Answer:
18
Explanation:
the numbers at the top tell you
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
1. 2650000
2. 2410
3. 48100