1,When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a soHowever, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble [Pb(OH)4]2–(aq) complex ion.lid precipitate forms.
2,The Ksp of strontium carbonate, SrCO3, is 5.60 × 10-10. Calculate the solubility of this compound in g/L.
3,Calculate the pH at the equivalence point for the titration of 0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of methylamine is 5.0× 10–4.
The first question makes no sense due to typos (and I think parts have been omitted).
The second one is
...........SrCO3 ==> Sr^2+ + CO3^2-
I..........solid.....0........0
C..........solid.....x........x
E..........solid.....x........x
Substitute the E line into Ksp expression and solve for x = solubility SrCO3.
#3. The pH will be determined by the hydrolysis of the salt which will be methylamine hydrochloride. And the concn of the salt will be 0.240/2 = 0.120M.
....CH3NH3H^+ + H2O ==> CH3NH2 + H3O^+
I.....0.120..............0........0
C......-x................x........x
E...0.120-x..............x.........x
Ka for CH3NH2*HCl = (Kw/Kb for CH3NH2) = (x)(x)/(0.120-x)
Solve for x = (H3O^+) and convert to pH.
1. When aqueous sodium hydroxide (NaOH) is added to a solution containing lead(II) nitrate (Pb(NO3)2), a solid precipitate of lead(II) hydroxide (Pb(OH)2) forms. The chemical equation for this reaction is:
Pb(NO3)2(aq) + 2NaOH(aq) -> Pb(OH)2(s) + 2NaNO3(aq)
To confirm that a solid precipitate of Pb(OH)2 forms, you can check the solubility rules. The solubility rule states that most hydroxide compounds are insoluble, except when they come from Group 1 metals (like NaOH) or ammonium (NH4+).
2. To calculate the solubility of strontium carbonate (SrCO3) in g/L, we need to use the solubility product constant (Ksp). The Ksp expression for SrCO3 is:
Ksp = [Sr2+][CO32-]
We can find the solubility (x) in g/L by assuming that the dissociation of SrCO3 is complete. So, the concentration of Sr2+ in the solution will be equal to the solubility (x), and the concentration of CO32- will be equal to 2x (since the stoichiometry is 1:2).
Substituting these values into the Ksp expression, we get:
Ksp = x * (2x)^2
5.60 x 10^-10 = 4x^3
Solving for x, we find:
x = (5.60 x 10^-10 / 4)^1/3
x ≈ 1.12 x 10^-3
Therefore, the solubility of SrCO3 in g/L is approximately 1.12 x 10^-3 g/L.
3. To calculate the pH at the equivalence point for the titration of 0.240 M methylamine (CH3NH2) with 0.240 M HCl, we need to consider the reaction that occurs between methylamine and HCl:
CH3NH2 + HCl -> CH3NH3+ + Cl-
Methylamine (CH3NH2) is a weak base, and HCl is a strong acid. Therefore, at the equivalence point, all the methylamine will be neutralized, and the solution will contain only the conjugate acid, CH3NH3+.
To calculate the pH at the equivalence point, we need to use the equilibrium expression for the ionization of methylamine, which is given by the base ionization constant (Kb):
Kb = [CH3NH3+][OH-] / [CH3NH2]
Since we have equal concentrations of methylamine and hydrochloric acid (0.240 M), at the equivalence point, the concentration of CH3NH3+ will also be 0.240 M.
Using the given Kb value of 5.0 x 10^-4, we can rearrange the equilibrium expression to solve for the concentration of hydroxide ions (OH-):
[OH-] = Kb * [CH3NH2] / [CH3NH3+]
[OH-] = (5.0 x 10^-4) * (0.240 M) / (0.240 M)
[OH-] = 5.0 x 10^-4
Since the solution contains both HCl (a strong acid) and CH3NH3+ (a weak acid), the pH at the equivalence point will be less than 7, but not extremely acidic. Therefore, the pH at the equivalence point is approximately 6.3.