1) How much water must be added to 500.mL of 0.200 M HCI to produce a 0.150M solution?
1 answer:
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Answer:
ksp = 0,176
Explanation:
The borax (Na₂borate) in water is in equilibrium, thus:
Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)
<em>When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ </em><em>(1)</em>
The ksp is defined as:
<em>ksp = [borate²⁻] [Na⁺]²</em>
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× = <em>3,00x10⁻³ mol of B₄O₇²⁻</em>
For (1), moles of Na⁺ are <em>3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺</em>
The [borate²⁻] is <em>3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = </em><em>0,353M</em>
And [Na⁺] is <em>6,00x10⁻³ mol of Na⁺ / 0,00850L = </em>0,706M
Replacing in the expression of ksp:
ksp = [0,353] [0,706]²
<em>ksp = 0,176</em>
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I hope it helps!
Answer:
Explanation:
The first step in this reaction is the<u> converstion from Kg</u> of <u>to</u> <u>grams</u> of
.
Then we can calculate the <u>percentage</u> of in the sample: