CALCULATE THE pH AND HYDROLYSIS IN 0.10 M AlCl3 SOLUTION.Ka = 1.2 X 10-5 FOR [Al(OH2)6]+3,OFTEN ABBREVIATED Al+3.TELL WHETHER pH IS ACID OR BASE.GIVE IONIC EQUATION.

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I really wanted more help than that. I know you can't solve it; if you could, you wouldn't have asked for help. I wanted to know WHAT about the problem you don't understand. Here are some hints to get you started.
[Al(H2O)6]3+ + H20 ==> [Al(H2O)5(OH)]2+ + H3O+

Write the Ka expression. If you let x = H3O^+, then the other product = x also, and the problem tells you that the starting material is 0.1 M. Plug in those values, solve for x and convert that to pH.
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To calculate the pH and hydrolysis in a 0.10 M AlCl3 solution, we can start by writing the ionization reaction:

[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+

The Ka expression for this reaction would be:

Ka = [Al(H2O)5(OH)]2+ / [Al(H2O)6]3+ [H3O+]

Since we're given that the concentration of the AlCl3 solution is 0.10 M, we can assume that the initial concentration of [Al(H2O)6]3+ is also 0.10 M. Let's assign 'x' as the concentration of [Al(H2O)5(OH)]2+ and [H3O+]:

Ka = x / (0.10 - x) x

Now we can solve for 'x'. Rearranging the equation and plugging in the Ka value (1.2 x 10^-5):

1.2 x 10^-5 = x / (0.10 - x) x

Solving this equation will give us the value of 'x', which represents the concentration of [Al(H2O)5(OH)]2+ and [H3O+]. Once we have that, we can find the pH using the formula:

pH = -log[H3O+]

Now, since you're looking for whether the pH is acidic or basic, we can determine that by looking at the concentration of [H3O+]. If [H3O+] is greater than 1 x 10^-7 M, the solution is acidic. If [H3O+] is less than 1 x 10^-7 M, the solution is basic.

Lastly, the ionic equation for the reaction is:

[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+

I hope this helps you solve the problem! Let me know if you have any further questions.

To calculate the pH and hydrolysis in a 0.10 M AlCl3 solution, we need to consider the hydrolysis reaction of the Al3+ ion with water.

The hydrolysis reaction for [Al(H2O)6]3+ can be represented as:
[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+

The equilibrium constant, Ka, is given as 1.2 x 10^-5.

To solve for the pH, we can use the following steps:

1. Write the expression for the equilibrium constant (Ka):
Ka = [Al(H2O)5(OH)]2+ * [H3O+] / [Al(H2O)6]3+

2. Since the concentration of [Al(H2O)6]3+ is provided as 0.10 M, we can represent it as [Al(H2O)6]3+ = 0.10.

3. Assume that the concentration of [Al(H2O)5(OH)]2+ and H3O+ at equilibrium is x.

4. Substitute the values into the Ka expression:
1.2 x 10^-5 = (x) * (x) / (0.10)

5. Solve the quadratic equation for x. We can assume that x << 0.10 since x represents the concentration of a weak acid. Therefore, we can simplify the equation as follows:
1.2 x 10^-5 = x^2 / 0.10

6. Solving for x:
x^2 = 1.2 x 10^-6
x = √(1.2 x 10^-6) = 3.464 x 10^-4 M

7. Calculate the pH using the concentration of H3O+:
pH = -log[H3O+]
pH = -log(3.464 x 10^-4)
pH ≈ 3.46

Therefore, the pH of the solution is approximately 3.46, indicating an acidic solution.

To write the ionic equation, we can represent the reaction as:

[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+

Please note that the actual concentration values may differ slightly due to the approximation made during the calculation.

To calculate the pH and hydrolysis in a 0.10 M AlCl3 solution, we need to consider the hydrolysis reaction of AlCl3 in water. The hydrolysis reaction can be represented as follows:

[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+

The equilibrium constant expression for this reaction is given by the Ka expression:

Ka = [Al(H2O)5(OH)]2+[H3O+]/[Al(H2O)6]3+

Given the value of Ka = 1.2 x 10^-5, we can proceed with solving the problem using the following steps:

1. Let x be the concentration of [H3O+] = [Al(H2O)5(OH)]2+ in moles per liter.
2. At equilibrium, the concentration of [Al(H2O)6]3+ = 0.10 - x, as the Al3+ ion comes from AlCl3, which is initially 0.10 M.
3. Substitute the concentration values into the Ka expression:

(1.2 x 10^-5) = [Al(H2O)5(OH)]2+[H3O+]/[Al(H2O)6]3+

(1.2 x 10^-5) = x^2/(0.10 - x)

4. Solve the quadratic equation for x. Rearrange the equation:

x^2 = (1.2 x 10^-5)(0.10 - x)

x^2 = 1.2 x 10^-6 - (1.2 x 10^-5)x

x^2 + (1.2 x 10^-5)x - 1.2 x 10^-6 = 0

5. Solve the quadratic equation using the quadratic formula or factoring.

x = (-b ± √(b^2 - 4ac))/2a

where a = 1, b = 1.2 x 10^-5, and c = -1.2 x 10^-6.

6. Once you have the value of x, convert it to pH using the equation pH = -log[H3O+].

To determine whether the pH is acidic or basic, compare the pH value to 7. If the pH is less than 7, the solution is acidic, and if the pH is greater than 7, the solution is basic.

To provide the ionic equation, we can rewrite the hydrolysis reaction in its ionic form:

[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+
Al3+ + 6H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+

pH IS 2.96