The four ionic species initially in solution are Na⁺, PO₄³⁻, Cr³⁺, and Cl⁻. Since the precipitate is composed of Cr³⁺ and PO₄³⁻ ions, the spectator ions must be Na⁺ and Cl⁻.
The complete ionic equation is 3Na⁺(aq) + PO₄³⁻(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → 3Na⁺(aq) + 3Cl⁻(aq) + CrPO₄(s).
So the balanced <u>net ionic equation</u> for this reaction would be Cr³⁺(aq) + PO₄³⁻(aq) → CrPO₄(s).
Carbon has the highest ionization energy.
the reaction is HF + NaOH which leads to NaF
Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4

Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of
and
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:


Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M