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spayn [35]
3 years ago
5

What is the final acetate ion concentration when 69. g Ba(C2H3O2)2 is dissolved in enough water to make 970. mL of solution? The

molar mass of Ba(C2H3O2)2 is 255.415 g/mol.
0.28 M
0.64 M
0.91 M
0.20 M
0.56 M
Chemistry
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:

<em><u></u></em>

  • <em><u>0.56M</u></em>

Explanation:

<u />

<u>1. Dissociation equation</u>

<u />

Assuming 100% dissociation, the equation is:

  • Ba(C₂H₃O₂)₂    →   Ba²⁺   +   2C₂H₃O₂⁻

                                                          ↑

                                                    acetate ions

<u>2. Molarity</u>

<u />

Calculate the molarity, M, of the solution:

  • M = n / V(in liters)

  • n = mass in grams / molar mass

  • n = 69.g / 255.415g/mol = 0.27015 mol

  • M = 0.27015mol / 0.970liter = 0.27850 mol/liter ≈ 0.28M

<u>3. Acetate ions</u>

From the chemical equation, 1 mol of dissolved Ba(C₂H₃O₂)₂ produces 2 acetate ions in solution.

Thus, 0.28 mol/liter × 2 = 0.56 mol/liter = 0.56M ← answer

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A 25.00-mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of
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0.0611M of HNO3

Explanation:

<em>The concentration of the NaOH solution must be 0.1198M</em>

<em />

The reaction of NaOH with HNO3 is:

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<em>1 mole of NaOH reacts per mole of HNO3.</em>

That means the moles of NaOH used in the titration are equal to moles of HNO3.

<em>Moles HNO3:</em>

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7 0
3 years ago
The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc
jek_recluse [69]

Answer: K_{sp}=1.25\times 10^{-14}

pK_{sp}=13.90

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as K_{sp}  

The equation for the ionization of magnesium phosphate is given as:

Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}

 When the solubility of Mg_3(PO_4)_2 is S moles/liter, then the solubility of Mg^{2+} will be 3S moles\liter and solubility of PO_4^{3-} will be 2S moles/liter.

Thus S = 0.173 g/L or \frac{0.173g/L}{262.8g/mol}=0.00065mol/L

K_{sp}=(3S)^3\times (2S)^2

K_{sp}=108S^5

K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}

pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90

5 0
3 years ago
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