Question

A student 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.

Answer 1

ksp = 0,176

Step-by-step 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×
`(1molB_(4)O_(7)^(2-))/(2molHCl)` = 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!