When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M
The volume of the gas will be decreased. Answer lies on the understanding of kinetic energy of particles and how particles occupy certain amount of space.
<span>I forgot to add: the answer is 50.91 difference in temperature. </span><span><span>
</span></span>
Because the size of atoms increase as you move down the periodic table.
<u>Answer:</u> The solubility product of silver (I) phosphate is ![9.57\times 10^{-10}](https://tex.z-dn.net/?f=9.57%5Ctimes%2010%5E%7B-10%7D)
<u>Explanation:</u>
We are given:
Solubility of silver (I) phosphate = 1.02 g/L
To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:
Molar mass of silver (I) phosphate = 418.6 g/mol
![\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L](https://tex.z-dn.net/?f=%5Ctext%7BMolar%20solubility%20of%20silver%20%28I%29%20phosphate%7D%3D%5Cfrac%7B1.02g%2FL%7D%7B418.6g%2Fmol%7D%3D2.44%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of silver (I) phosphate follows:
3s s
The expression of
for above equation follows:
![K_{sp}=(3s)^3\times s](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%283s%29%5E3%5Ctimes%20s)
We are given:
![s=2.44\times 10^{-3}M](https://tex.z-dn.net/?f=s%3D2.44%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above expression, we get:
![K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%283%5Ctimes%202.44%5Ctimes%2010%5E%7B-3%7D%29%5E3%5Ctimes%20%282.44%5Ctimes%2010%5E%7B-3%7D%29%5C%5C%5C%5CK_%7Bsp%7D%3D9.57%5Ctimes%2010%5E%7B-10%7D)
Hence, the solubility product of silver (I) phosphate is ![9.57\times 10^{-10}](https://tex.z-dn.net/?f=9.57%5Ctimes%2010%5E%7B-10%7D)