1. calculate H+ ion concentrates in 0.10 m acetic acid solutions. Given that dissociation constant of acetic acid in water is 1.8 * 10^-5 ?

2. Calculate the concentration of H+ and OH- in a 0.05 m HNO3 solutions at 25^0c ?
3. The degree of dissociation of pure water at 18^0c is found to be 1.8 * 10^-9. Find the ionic product of water and its dissociation constant at 18^0c ?

1. If that is 0.1 molal, you must change molality to molarity (but I don't see a density listed) [or you can assume 0.1 m = 0.1 M and not make a significant difference).

CH3COOH ==> CH3COO^- + H^+

Set up an ICE chart, substitute into the expression for Ka = (H^+)(CH3COO^-)/(CH3COOH), and solve for (H^+)

2. HNO3 is a strong acid; therefore, H^+ = M

3. Here is a site you can read about degree of dissociation. Use that to calculate concn of ions, convert to the ion product and Kion from there.
http://www.pinkmonkey.com/studyguides/subjects/chem/chap12/c1212601.asp
Post your work if you get stuck.

1. To calculate the H+ ion concentration in a solution of acetic acid, you need to use the dissociation constant (Ka) and the initial concentration of the acid. The dissociation constant of acetic acid (CH3COOH) in water is given as 1.8 * 10^-5.

The dissociation equation for acetic acid (CH3COOH) is:
CH3COOH ⇌ CH3COO- + H+

Let's assume the initial concentration of acetic acid is 0.10 M.

To calculate the concentration of H+ ions, we will first calculate the concentration of CH3COO- ions at equilibrium using the expression for equilibrium constant (Ka):
Ka = [CH3COO-][H+] / [CH3COOH]

Since the initial concentration of CH3COOH is 0.10 M and the degree of dissociation (α) for acetic acid is unknown, we can say that the concentration of CH3COOH that dissociates is α * 0.10 M, and the concentration of CH3COO- and H+ ions formed is α * 0.10 M.

Therefore, the expression for Ka becomes:
1.8 * 10^-5 = (α * 0.10)(α * 0.10) / (0.10 - α * 0.10)

Now you can rearrange this equation to solve for α, which represents the degree of dissociation of acetic acid. Once you find the value of α, you can use it to calculate the concentration of H+ ions.

2. To calculate the concentration of H+ and OH- ions in a solution of HNO3, you can use the concept of strong acids. HNO3 is a strong acid, which means it dissociates completely in water, forming H+ and NO3- ions.

In a 0.05 M HNO3 solution, the H+ ion concentration will be equal to the initial concentration of the acid, i.e., 0.05 M.

As for the OH- ions, the concentration will be negligible because HNO3 is an acid and doesn't contribute significantly to the OH- concentration.

3. The degree of dissociation (α) of pure water at 18°C is given as 1.8 * 10^-9. The degree of dissociation represents the fraction of water molecules that have dissociated into H+ and OH- ions.

The concentration of H+ and OH- ions in pure water at 18°C will be equal to the degree of dissociation (α) times the initial concentration of water molecules. Since water has a concentration of 1 M, the concentration of H+ and OH- ions will be α * 1 M.

To find the ionic product of water (Kw), you need to multiply the concentrations of H+ and OH- ions. In this case, Kw = [H+][OH-] = (α * 1)(α * 1) = α^2.

The dissociation constant (Kw) at 18°C will be the same as the ionic product, which is α^2.