Astudent who is performing this experiment pours an 8.50 ml sample of the saturated borax solution into a 10 ml graduated cylinder after the borax solution had cooled to a certain temperature t. the student rinses the sample into a small beaker using distilled water, and then titrates the solution with a 0.500 m hcl solution. 12.00 ml of the hcl solution is needed to reach the endpoint of the titration.
calculate the value of ksp for borax at temperature t.

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ (1)

The ksp is defined as:

ksp = [borate²⁻] [Na⁺]²

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl× = 3,00x10⁻³ mol of B₄O₇²⁻

For (1), moles of Na⁺ are 3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺

The [borate²⁻] is 3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = 0,353M

And [Na⁺] is 6,00x10⁻³ mol of Na⁺ / 0,00850L = 0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

ksp = 0,176

I hope it helps!

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explanation: