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olya-2409 [2.1K]
3 years ago
5

A small crystal of the slightly soluble salt, CaSO4, dissolves in a solution of calcium sulfate. The original solution must have

been:
dilute and saturated
dilute and unsaturated
concentrated and saturated
concentrated and unsaturated
Chemistry
2 answers:
Margarita [4]3 years ago
5 0

Answer:

dilute and unsaturated

Explanation:

A dilute solution has a low concentration of solute in solvent. A concentrated solution has a high concentration of solute in solvent. In an unsaturated solution more solute can be dissolved in the solvent to form a solution. In a saturated solution more solute cannot be dissolved in the solvent to form a solution.

Since more CaSO₄ can be dissolved in CaSO₄ solution the solution is both dilute and unsaturated.  

xz_007 [3.2K]3 years ago
3 0
It has to be dilute since the crystal dissolved and unsaturated for the same reason so B
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<u>Answer:</u> The pH of acid solution is 4.58

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To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

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Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

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Molarity of propanoic acid solution = 0.6100 M

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Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

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