Question

Part a what is the concentration of k+ in 0.15 m of k2s? express your answer to one decimal place and include the appropriate units. 0.3 m submithintsmy answersgive upreview part correct part b if cacl2 is dissolved in water, what can be said about the concentration of the ca2+ ion? if is dissolved in water, what can be said about the concentration of the ion? it has the same concentration as the cl− ion. its concentration is half that of the cl− ion. its concentration is twice that of the cl− ion. its concentration is one-third that of the cl− ion. submithintsmy answersgive upreview part correct part c a scientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. how many grams of na3po4 will be needed to produce 650. ml of a solution that has a concentration of na+ ions of 1.30 m ?

Answer 1

1) Answer is: the concentration of potassium cations (K⁺) is 0.3 M.

Balanced chemical reaction (dissociation) of potassium sulfide in the water:

K₂S(aq) → 2K⁺(aq) + S²⁻(aq).

c(K₂S) = 0.15 M.

From balanced reaction: n(K₂S) : n(K⁺) = 1 : 2, because volume of the solution is constant: c(K⁺) = 2 · c(K₂S).

c(K⁺) = 0.3 M; concentration of potassium cations.

2) Answer is: its concentration is half that of the Cl⁻ ion.

Balanced chemical reaction (dissociation) of calcium chloride (CaCl₂) in the water:

CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq).

From balanced reaction: n(Ca²⁺) : n(Cl⁻) = 1 : 2, because volume of the solution is constant: c(Ca²⁺) = c(Cl⁻) ÷ 2.

n is amount of the substance.

3) Answer is: 46.176 grams of Na₃PO₄ will be needed.

Balanced chemical reaction (dissociation) of sodium phosphate  (Na₃PO₄) in the water:

Na₃PO₄(aq) → 3Na⁺(aq) + PO₄³⁻(aq).

V = 650 mL ÷ 1000 mL.

V = 0.650 L; volume of the solution.

c(Na⁺) = 1.3 M; concentratium of sodium cations.

From balanced reaction: n(Na₃PO₄) : n(Na⁺) = 1 : 3, because volume of the solution is constant: c(Na₃PO₄) = c(Na⁺) ÷ 3.

c(Na₃PO₄) = 1.3 M ÷ 3.

c(Na₃PO₄) = 0.43 M.

n(Na₃PO₄) = c(Na₃PO₄) · V(solution).

n(Na₃PO₄) = 0.43 M · 0.65 L.

n(Na₃PO₄) = 0.281 mol.

m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).

m(Na₃PO₄) = 0.281 mol · 163.94 g/mol.

m(Na₃PO₄) = 46.176 g.

Answer 2

1. Concentration K⁺ = [K⁺] = 0.3 M

2. Ca²⁺concentration is half that of the Cl⁻ ion.

3. 45,838 grams of Na₃PO₄ will be needed

Further explanation

Stokiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.

Reaction equations are chemical formulas for reagents and product substances

Reaction coefficients are numbers in the chemical formula of substances involved in the reaction equation. Reaction coefficients are useful for balancing reagents and products.

The reaction coefficient shows the ratio of the number of moles or molecules of the reacting substance

The ionization reaction is the reaction of the decomposition of a substance into its ions when the substance is dissolved in water.

Molarity (M)

Molarity shows the number of moles of solute in every 1 liter of solution.

$\large {\boxed{\bold {M ~ = ~ \frac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = volume of the solution

• 1. The decomposition / ionization reaction of K₂S is

K₂S(aq) ----> 2K⁺ (aq) + S²⁻ (aq)

Comparison of reaction coefficients = mole ratio = concentration ratio for the same volume, so

K₂S coefficient: coefficient 2K⁺ = 1: 2

[K⁺] = 2. [K₂S]

[K⁺] = 2. 0.15 M

[K⁺] = 0.3 M

• 2. The decomposition / ionization reaction of CaCl₂ is

CaCl₂ -> Ca²⁺ + 2Cl⁻

Same as the explanation in point 1:

coefficient Ca²⁺: coefficient Cl⁻= 1: 2

so that

 [Ca²⁺] = 1/2 [Cl⁻]

[Ca²⁺] half of the concentration [Cl⁻]

• 3. Decomposition / ionization reactions from Na₃PO₄

Na₃PO₄ ---> 3Na + + PO₄³⁻

[Na₃PO₄] = 1/3. [Na +]

[Na₃PO₄] = 1/3. 1.3

[Na₃PO₄] = 0.43 M

then the number of moles is

mol = M. vol

mole = 0.43. 0.65 L

mole = 0.2795

 So that its mass:

molar mass of Na₃PO₄ = 164 g / mol

mass = mole. molar mass

mass = 0.2795. 164

mass = 45,838 grams

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Keywords: ionization, molar mass, mole, mass, Na₃PO₄, K₂S, CaCl₂