Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
You need to know the energy frequency relationship for photons, which is thanks to Max Planck:
Photon Energy = Planck constant x Frequency
Rarranged:
Photon Energy / Planck Constant = Frequency
Planck Constant = 6.63x10^-34
2.93x10^-25 / 6.63x10^-34 = Frequency
Answer:
Base Mg(OH)2 does neutralise the acid and is 12g in excess.
Explanation:
2HCL +Mg(OH)2 -> MgCl2 + 2H20
2 * 36.458 g of HCL react with 58.319 g of Mg(OH)2 to neutralise it.
72.916 HCl reacts with 58.319 g of the base.
So 20 g HCl reacts with (58.319/72.916) * 20 = 16g.
There are 28 g of Mg(OH)2 so the base does neutralise all the acid.
The Mg(OH)2 is 28 - 16 = 12 g in excess.
