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a 10 ml sample of borax- borate equilibrium solution reacts completely with 23.33 ml of a 0.1006 M HCl solution. calculate the concentration of borate and then the Ksp for borax​


Sagot :

Answer:

See answer below

Explanation:

To do this, you need to write down the chemical reaction giving place here.

As this is to explain the thermodynamics of the borax-borate reaction, the general reaction is as follow:

Na₂B₄O₅(OH)₄.8H₂O(s) <---------> 2Na⁺(aq) + B₄O₅(OH)₄²⁻(aq) + 8H₂O(l)

Where:

Na₂B₄O₅(OH)₄.8H₂O(s) = Borax

B₄O₅(OH)₄²⁻(aq) = Borate ion

Now, in order to calculate the concentration of borate, this compound undergoes a titration with HCl. The reaction that takes place here is:

B₄O₅(OH)₄²⁻(aq) + 2HCl + 3H₂O <---------> 4H₃BO₃(aq) + 2Cl⁻(aq)

According to this reaction, it takes 2 moles of HCl to completely react with the borate ion, therefore, the expression to calculate the concentration will be:

2 nHCl = nB₄O₅(OH)₄²⁻

2M₁V₁ = M₂V₂

Replacing the data of concentration and volume of both compounds we have:

2 * 0.1006 * 23.33 = 10M₂

M₂ = 2 * 0.1006 * 23.33 / 10

M₂ = 0.4693 M

This is the concentration of the borate. Now, to calculate the Ksp we need to write the equilibrium reaction taking place with the borax:

Na₂B₄O₅(OH)₄.8H₂O(s) <---------> 2Na⁺(aq) + B₄O₅(OH)₄²⁻(aq) + 8H₂O(l)

Ksp = [B₄O₅(OH)₄²⁻] [Na⁺]²

As we already have the concentration of borax, we can also know the concentration of sodium. If we call concentration as "s" (Stands for solubility), and rewrite the expression of Ksp:

Ksp = s * (2s)²

Finally, replacing the value of the concentration we have:

Ksp = (0.4693) (2*0.4693)²

Ksp = 0.4693 * 0.881

Ksp = 0.4134

Hope this helps