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Nina [5.8K]
3 years ago
8

What is the molar solubility of calcium carbonate ( CaCO3 ) in water? The solubility-product constant for CaCO3 is 4.5 × 10-9 at

25 °C.
A) 8.35
B) 9.5 × 10-5
C) 9.0 × 10-9
D) 2.3 × 10-9
E) 6.7 × 10-5
Chemistry
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

Option E)

  • 6.7\times 10^{-5}M

Explanation:

<u />

<u>1. Write the solubility equation:</u>

  • CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻

<u>2. Write the concentrations below the equation:</u>

  • CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻

             A - s               s            s

<u>3. Write the Ksp equation:</u>

  • Ksp = [Ca²⁺] . [CO₃²⁻]       ↔ the solid substances do not appear
  • Ksp = s × s
  • 4.5 × 10⁻⁹ = s²

<u>4. Solve the equation:</u>

<u />

       s=\sqrt{4.5\times 10^{-9}M^2}=6.7\times 10^{-5}M

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2 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

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Molarity of NaOH = 0.150 M

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Step 3: Calculate the number of moles hydrocyanic acid (HCN)

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Step 4: Calculate the limiting reactant

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HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

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//

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