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Ganezh [65]
3 years ago
12

Calcium hydroxide is partially soluble as shown by the

Chemistry
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

4

10

Explanation:

The reaction equation is given as;

            Ca(OH)₂     →    Ca²⁺   + 2OH⁻

Concentration of Ca(OH)₂ = 5 x 10⁻⁵M

         

Unknown:

pOH of the solution  = ?

pH of the solution = ?

Solution:

Solve for the pOH of this solution using the expression below obtained from the ionic product of water;

                 pOH  = ⁻log₁₀[OH⁻]

           Ca(OH)₂             →           Ca²⁺              +              2OH⁻

          1moldm⁻³                         1moldm⁻³                     2 x 1moldm⁻³

         5 x 10⁻⁵moldm⁻³         5 x 10⁻⁵moldm⁻³            2( 5 x 10⁻⁵moldm⁻³ )

                                                                                            1 x 10⁻⁴moldm⁻³

  Therefore;

             pOH = -log₁₀ 1 x 10⁻⁴ = 4

Since

           pOH + pH  = 14

                        pH  = 14  - 4  = 10

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8 0
3 years ago
Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of Ka, pKa=−logKa The Henderson-Hasselbalch equati
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Answer:

You need to add 109,2g of NH₄Cl.

Explanation:

To calculate the pH in a buffer you can use Henderson-Hasselbalch formula:

pH = pka + log₁₀\frac{[base]}{[acid]}

<em>1.</em> ka of 1,8x10⁻⁵ ≡ 4,74

pH = 4,74 +  log₁₀\frac{[1]}{[10]} = <em>3,74</em>

pH = 4,74 +  log₁₀\frac{[1]}{[1]} = <em>4,74</em>

pH = 4,74 +  log₁₀\frac{[10]}{[1]} = <em>5,74</em>

<em>2. </em>Using:

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The moles of ammonia (base) are:

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Replacing:

2,32 = \frac{[acid]}{[0,88]}

[acid] = 2,0416 moles of NH₄Cl ₓ (53,491g/mol) = <em>109,2 g of NH₄Cl</em>

You need to add 109,2g of NH₄Cl.

I hope it helps!

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