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Brut [27]
3 years ago
5

2. 25.0mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small me

asured amount of Mg2 requires 2.12mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2 is added requires 25.88mL of the EDTA to reach the end point. a. How many mL of EDTA are needed to titrate the Ca2 ion in the aliquot
Chemistry
1 answer:
LekaFEV [45]3 years ago
5 0

Answer:

2) 25.0mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg2+ requires 2.12mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg2+ is added requires 25.88mL of the EDTA to reach the end point.

a. How many mL of EDTA are needed to titrate the Ca2+ ion in the aliquot?

b. How many moles of EDTA are there in the volume obtained in a.?

c. What is the molarity of the EDTA solution?

Explanation:

Given that;

Volume of aliquot = 25mL

Blank reading = 2.12mL

2a)

Volume of EDTA used for Ca²⁺ ion

25.88mL - 2.12mL = 23.76mL

Therefore mL of EDTA needed to titrate the Ca²⁺ ion in the aliquot is 23.76mL

2b)

Molarity of Ca²⁺ ion  is 0.0172M

Mole of EDTA =

=\frac{0.0172 moles}{1l} \times (25mL) \times \frac{1L}{1000mL} \\\\=0.00043moles

2c)

Molarity of EDTA = mole of EDTA / Vol. of EDTA

=(\frac{0.00043mole}{23.76mL} )\times \frac{1000mL}{1L} \\\\=0.0180M

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(1) The solubility of Salt AB2(S) IS 5mol/dm^3.
Svetlanka [38]

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

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1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

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