The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:
PbI2 = Pb^2+ + 2I-
We solve as follows:
Ksp = [Pb2+][I-]^2 = <span>1.4 x 10-8
</span><span>1.4 x 10-8 = x(2x)^2
</span><span>1.4 x 10-8 = 4x^3
x = 1.5x10^-3 M
The molar solubility would be </span>1.5x10^-3 M.
Answer:
2Mg (s) + 
(g) ---> 2MgO (s)
Explanation:
Study your experiment setup.<span> In 30 minutes, how will the air temperature in the bottles compare?</span><span> What do you predict will happen to the ice in each bottle?</span>
Answer:
a) AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Explanation:
a) AgNO3 + KI → Ag+ + NO3- + K+ + I-
Ag+ + NO3- + K+ + I- → AgI + KNO3
AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-
Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-
6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3
2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-
2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
