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Korolek [52]
3 years ago
7

A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

dded so that the following salts are formed. Salt BaC2O4. ZnC2O4. Ag2C2O4.
Ksp. 1.5 × 10−8. 1.35 × 10−9. 1.1 × 10−11
What is the order of the cations that would precipitate with the addition of C2O42−?

Every time I do this I keep getting the opposite of the answer. Can you please show your process so I can see where I am going wrong? Thank you!
Chemistry
1 answer:
Jet001 [13]3 years ago
4 0

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

                 BaC₂O₄ ⇌   Ba²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

                ZnC₂O₄ ⇌   Zn²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

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alexandr402 [8]

Answer:

LiCl - pink

NaCl - yellow

KCl - purple/blue-violet

CaCl₂ - orange/orange-red

Hope this helps.

4 0
3 years ago
What mass is 8.9 moles of Ca?<br> 0.22<br> mole(s)<br> 8.9 moles of Ca x<br> grams<br> 8.9<br> grams
iren2701 [21]
Can you restate the questions again ?
4 0
3 years ago
A spectator ion is (Select all that apply.) a. an ionic component of a reactant that is unchanged by the reaction your eye, b. c
DENIUS [597]

Correct Question:

A spectator ion is (Select all that apply.)

- a piece of french fry contaminating the reaction mixture

- an ionic component of a reactant that is unchanged by the reaction

-in this experiment, nitrate ion

- your eye, carefully watching the progress of the reaction

Answer:

- an ionic component of a reactant that is unchanged by the reaction

Explanation:

A spectator ion is an ion that exists as a reactant and a product in a chemical equation. A spectator ion is one that exists in the same form on both the reactant and product sides of a chemical reaction.

Spectator ions are ions that are present in a solution but don't take part in the reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. The other ions don't take part in this chemical reaction and are therefore called spectator ions.

The correct option is therefore the option;

- an ionic component of a reactant that is unchanged by the reaction

5 0
3 years ago
How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----&gt; K3PO4 + 3
8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

\frac{3}{1} =\frac{y}{0.02256} \\

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

5 0
3 years ago
A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
Ludmilka [50]

Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) <em>Moles of glucose </em>

n = 10.20 g × (1 mol/180.16 g)

  = 0.056 62 mol

(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

3 0
3 years ago
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