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4vir4ik [10]
3 years ago
11

Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360

g/L at room temperature (25 °C).
How can Yasmin make a supersaturated saline solution?


She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature (25 °C).

She can add 360 g of salt to 1 L of room temperature water (25 °C) and stir the solution until all the salt dissolves.

She can add 380 g of salt to 1 L of cold water (5 °C) and stir the solution until most of the salt dissolves. Then, she can carefully heat the solution to room temperature (25 °C).

She can add 380 g of salt to 1 L of room temperature water (25 °C) and let the solution sit for 24 hours, so the salt dissolves.
Chemistry
2 answers:
KatRina [158]3 years ago
6 0

B just took test and got the correct answer.

Aleks [24]3 years ago
3 0

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.  

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

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The new pH is 7.69.

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The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

                       pH = pKa + log10 ([A–]/[HA])

Here, 100 mL  of  0.10 m TRIS buffer pH 8.3

                 pka = 8.3

         0.005 mol of TRIS.

∴  8.3 = 8.3 + log \frac{[0.005]}{[0.005]}

<em>    </em>inverse log 0 = \frac{[B]}{[A]}

   \frac{[B]}{[A]} = 1

Given; 3.0 ml of 1.0 m hcl.

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pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69

Therefore, the new pH is 7.69.

Learn more about pH here:

brainly.com/question/24595796

#SPJ1

 

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