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aksik [14]
2 years ago
5

When an ionic solid is added into a solvent, you can see that the ionic solid dissociates into its respective cations and anions

. The solvent molecules immediately cluster around the ions. This is known as solvation. When the solvent is water, the solvation process is called hydration. Dissolution and hydration occur simultaneously, and salt is said to be dissolved in water. You can observe that some of the ions dissolved in the solution recrystallize or deposit to form the solid salt. When the rate of dissolution equals the rate of recrystallization (deposition), dynamic equilibrium is reached.
Chemistry
1 answer:
SVEN [57.7K]2 years ago
4 0

The order of the steps, from first to last, is as follows based on the information in the question:

1. Salt is broken down into cations and anions.

2. Anion hydration.

3. Cation hydration.

4. Cations and anions that have been dissolved start to settle as a solid salt.

5. The rate of dissolution and recrystallization are equal.

Without hydration of the ions, the cation and anion cannot get separated. The three processes of ion dissociation, cation hydration, and anion hydration must all take place at once.

<h3>Why water is an effective solvent?</h3>

Water is a great solvent that can dissolve a wide variety of compounds due to its polarity and capacity to create hydrogen bonds.

Learn more about solvation here:

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Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH
uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

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umka2103 [35]
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Explanation:

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