Chemistry help please? ?

1.
All sodium salts (such as NaCl), all nitrates (such as AgNO₃), all ammonium salts (such as NH₄Br) and all actates (such as Pb(C₂H₃O₂)₂) are soluble in water. PbI₂ is insoluble.

The answer: c. PbI₂

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2.
Mg₃(PO₄)₂(s) ⇌ 3Mg²⁺(aq) + 2PO₄³⁻(aq) … Ksp = [Mg²⁺]³ [PO₄³⁻]²

The answer: a./e. Ksp = [Mg²⁺]³ [PO₄³⁻]²

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3.
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq) … Ksp = [Ca²⁺] [F⁻]²

Dissolving each mole of CaF₂ gives 1 mole of Ca²⁺ ions.
Hence, solubility = [Ca²⁺] = 2.15 × 10⁻⁴ M
[F⁻] = [Ca²⁺] × 2 = (2.15 × 10⁻⁴) × 2 = 4.3 × 10⁻⁴ M

Ksp = [Ca²⁺] [F⁻]² = (2.15 × 10⁻⁴) × (4.3 × 10⁻⁴)² = 3.98 × 10⁻¹¹

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4.
Ba(IO₃)₂(s) ⇌ Ba²⁺(aq) + 2IO₃⁻(aq) … Ksp = 1.5 × 10⁻⁹
Equilibrium: _ s M _____ 2s M

At equilibrium:
Ksp = [Ba²⁺] [IO₃⁻]²
9.8 × 10⁻⁹ = s × (2s)²
4s³ = 1.5 × 10⁻⁹
s = ³√(1.5 × 10⁻⁹ / 4) = 7.21 × 10⁻⁴
Solubility = 7.21× 10⁻⁴ M

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5.
NaI is soluble in water. Hence, initial [I⁻] = 0.2 M

__PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq) … Ksp = 9.8 × 10⁻⁹
Initial: ____ 0 M _____ 0.2 M
Change: __ +s M _____ + 2s
Eqm: _____ s M ___ (0.2 + 2s) M

As Ksp is very small, the solubility of PbI₂ is very small.
It can be assumed that 0.2 ≫ 2s
Equilibrium [I⁻] = (0.2 + 2s) M ≈ 0.2 M

At equilibrium:
Ksp = [Pb²⁺] [I⁻]²
9.8 × 10⁻⁹ = s × (0.2)²
s = (9.8 × 10⁻⁹) / (0.2)² = 2.45 × 10⁻⁷
Molar solubility = 2.45 × 10⁻⁷ M

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6.
Refer to: https://www.chm.uri.edu/weuler/chm112/refmater/KspTable.htm
Ksp for BaSO₄ = 1.1 × 10⁻¹⁰
(Ksp values from different sources may be slightly different.)

Volume of the final solution = (50.0 + 86.4) mL = 136.4 mL

When the two solutions are mixed, both of the solution is diluted.
[Ba²⁺] after mixing = (1.0 M) × (50/136.4) = 0.367 M
[SO₄²⁻] after mixing = (0.494 M) × (86.4/136.4) = 0.313 M

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) … Ksp = 1.1 × 10⁻¹⁰
Reaction quotient, Qc = 0.367 × 0.313 = 0.115 > Ksp
Hence, precipitate forms.

1
check a table of solubilities
https://en.wikipedia.org/wiki/Solubility_table
it's probably C

2
in general
AxBy ↔ X A(+) + Y B(-)
Ksp = [A(+)]^x [B(-)]^y
so
Ksp = [Mg+2]^3 [PO4-3]^2

3
for each CaF2 you get 1 Ca+2 and 2 F-
so [Ca+2] = 2.15E-4 and [F-] = 4.30E-4
Ksp = [Ca+2] [F-]^2
I get about 4E-11

4
let solubility be X
then [Ba+2]=X and [IO3-]=2X
set up Ksp and solve for X
I get about 4E-4 M


5
NaI supplies I- on a one to one basis
so
[Pb+2] = X and [I-] = 2X + 0.20
set up Ksp and solve for X
you may find this useful
https://www.wolframalpha.com/widgets/view.jsp?id=3f4366aeb9c157cf9a30c90693eafc55
I get about 2.5E-7 M

6
from a balanced equation you can see that BaSO4 might precipitate
so we need the conc of the ions
[Ba+2] = (0.0500L*1mol/L) / (0.0500+ 0.0864) L = 0.3666 M
[SO4-2] = (0.0864L* 0.494mol/L)  / (0.0500+ 0.0864) L = 0.3129 M
now do a trial run for Ksp
if the trial is grater than the given value of Ksp, the a precip will form
otherwise not
I got an ion procuct (trial value) of about 0.1 and looked up a Ksp for BaSO4 of 1.1E-10
my results predict a ppt.

When you get a good response,
please consider giving a best answer.
This is the only reward we get.

1. c. PbI2

2. I see no difference between answers a. and e., both of which are correct, assuming there is a minus sign behind the last "3".

3.  The solubility of CaF2 is NOT "2.15 × 10 4 M".  It might be 2.15 × 10^− 4 M, in which case:
Ksp = [2.15 x 10^-4] x [4.30 x 10^-4]^2 =  3.98 x 10^-11

4.
Let z be the molarity of Ba(IO3)2 to be found:
[z] x [2z]^2 = 1.5 × 10^–9
Solve for z algebraically:
z = 0.00072 M