<span>solution of KI becomes saturated at 10 degrees when around 135-138g KI are added to 100 g of water, so it should be still unsaturated, A. unsaturated (although it is very close to saturation)</span>
0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
I looked up what is the molecular formula for Phosphine and got this: PH3
Hope this helps! Let my know if this was correct.
