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Romashka [77]
3 years ago
14

A chemistry student is given 600. mL of a clear aqueous solution at 27.° C. He is told an unknown amount of a certain compound X

is dissolved in the solution. The student allows the solution to cool to 27.° C. The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 28.2 g.
Using only the information from above, can you calculate the solubility of X at 21.° C?
Chemistry
1 answer:
DiKsa [7]3 years ago
8 0

Answer: No

Explanation:

Firstly, the molar mass of the dissolved solid is not given. This is necessary to calculate the number of moles present in solution. Secondly, solubility always has to do with temperature and the specified temperature is 27°c and not 21°c. This makes it impossible to calculate the solubility at 21°c. Further information must supplied before the solubility at 21°c can be accurately calculated.

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A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what i
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Answer:

3.07 × 10⁻⁴

Explanation:

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pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M

Step 2: Calculate the concentration of HY

5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:

Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}

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