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White raven [17]
3 years ago
6

How many moles are there in 25 grams of copper (II) sulphate crystals, CuSO4.5H2O

Chemistry
1 answer:
kondaur [170]3 years ago
8 0
Moles = mass/molar mass of CuSO4.5H2O
 
           =    25/249.68
           =    0.1 moles.
Hope this helps!
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A certain weak acid, HA, has a Ka value of 6.7 X 10^-7.
yuradex [85]

Answer:

  • <u>Part A: 0.26% </u>

  • <u>Part B: 0.82%</u>

Explanation:

<em>Percent ionization</em> is the percent of the original acid that has ionized:

  • %, ionization = (molar concentration of hydrogen ions at equilibrium / molar concentration of original acid) × 100

<u><em>Part A:</em></u>

<u>1) Data:</u>

  • Ka: 6.7 × 10 ⁻⁷
  • [HA] = 0.10 M
  • %, ionization = ?

<u>2) Equilibrium equation:</u>

  • HA ⇄ H⁺ + A⁻

<u>3) ICE (initial, change, equilbirium) table </u>

                           Concentrations

                         HA            H⁺       A⁻

Initial                 0.10           0        0

Change             - x            + x     + x

Equilibrium     0.10 - x         x         x

  • Equation:      Ka          =         [H⁺] [A⁻] / [HA] =

                        6.7 × 10 ⁻⁷   =          x² / (0.10 - x)

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

Since Ka << 1, you can assume x << 0.10 and 0.10 - x ≈ 0.10

  • 6.7 × 10 ⁻⁷ ≈  x² / 0.10 ⇒ x² ≈ 6.7 × 10⁻⁸ ⇒ x ≈ 2.588 × 10⁻⁴

  • [H⁺] ≈ 2.588 × 10⁻⁴ M

  • % ionization ≈ (2.588 × 10⁻⁴ M / 0.1 M) × 100 ≈ 0.2588 % ≈ 0.26% (two significant figures)

<u><em>Part B:</em></u>

<u>1) Data:</u>

  • Ka: 6.7 × 10 ⁻⁷
  • [HA] = 0.010 M
  • %, ionization = ?

<u>2) Equilibrium equation:</u>

  • HA ⇄ H⁺ + A⁻

<u>3) ICE table:</u>

                            Concentrations

                           HA              H⁺       A⁻

Initial                 0.010            0         0

Change               - x              + x     + x

Equilibrium     0.010 - x           x         x

  • Equation:      Ka          =         [H⁺] [A⁻] / [HA] =

                        6.7 × 10 ⁻⁷   =          x² / (0.010 - x)

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

Since Ka << 1, you can assume x << 0.010 and 0.010 - x ≈ 0.010

  • 6.7 × 10 ⁻⁷ ≈  x² / 0.010 ⇒ x² ≈ 6.7 × 10⁻⁹ ⇒ x ≈ 8.185 × 10⁻5

  • [H⁺] ≈ 8.185 × 10⁻⁵ M

  • % ionization ≈ (8.185 × 10⁻⁵ M / 0.010 M) × 100 ≈ 0.8185 % ≈ 0.82% (two significant figures)
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