Ba(OH)2 contains 4.285gram of Ba (OH)2 in 100mL of solution the find molarity, PH,POH,[H^+] ?

To find the molarity of Ba(OH)2 in the solution, we first need to calculate the number of moles of Ba(OH)2 present in 4.285 grams.

The molar mass of Ba(OH)2 is:
Ba = 137.33 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

Molar mass of Ba(OH)2 = 137.33 + 2(16.00) + 2(1.01) = 171.35 g/mol

Number of moles of Ba(OH)2 = 4.285 g / 171.35 g/mol = 0.025 moles

Now, we need to calculate the molarity of Ba(OH)2 in the solution:

Molarity = moles of solute / volume of solution in liters
Molarity = 0.025 moles / 0.1 L = 0.25 M

Next, we can calculate the pOH of the solution:

pOH = -log[OH-]
pOH = -log(0.25) = 0.60

Since pH + pOH = 14, we can calculate the pH of the solution:

pH = 14 - pOH = 14 - 0.60 = 13.40

Finally, we can calculate the concentration of [H+] in the solution using the equation:

[H+] = Kw / [OH-]
[H+] = 1.0 x 10^-14 / 0.25 = 4.0 x 10^-14 M

Therefore, the molarity of Ba(OH)2 in the solution is 0.25 M, the pH is 13.40, the pOH is 0.60, and the concentration of [H+] is 4.0 x 10^-14 M.