I still need help with this problem...
Calculate the concentration of the solution
the molality of hydroxide ions in a solution prepared from 9.16 g of barium hydroxide dissolved in 179 g of water
I do not know how to approach this...I keep doing something wrong.
It's easy to post a question and say you're doing something wrong. You should post your work and let us see what you are doing wrong. We can catch hidden errors that way. Here is what you do.
Convert 9.16 g Ba(OH)2 to moles.
There are two OH^- per mole Ba(OH)2; therefore, moles Ba(OH)2 x 2 = moles OH^-
Then moles OH^-/kg water = m of OH^-
To calculate the concentration of the solution (molality of hydroxide ions), you need to follow these steps:
1. Convert the given mass of barium hydroxide (9.16 g) to moles. This can be done by dividing the mass by the molar mass of barium hydroxide (Ba(OH)2), which is calculated by adding the atomic masses of each element: Ba (137.33 g/mol), O (16.00 g/mol), and H (1.01 g/mol). The molar mass of Ba(OH)2 is 137.33 + (2 * 16.00) + (2 * 1.01) = 171.33 g/mol.
Moles of Ba(OH)2 = mass of Ba(OH)2 / molar mass of Ba(OH)2
= 9.16 g / 171.33 g/mol
2. Calculate the moles of hydroxide ions (OH-) since each barium hydroxide molecule dissociates to produce two hydroxide ions.
Moles of OH- = 2 * moles of Ba(OH)2
3. Calculate the mass of water in kilograms by dividing the given mass (179 g) by 1000.
Mass of water (in kg) = 179 g / 1000
4. Finally, compute the molality of hydroxide ions in the solution by dividing the moles of hydroxide ions by the mass of water in kilograms.
Molality = Moles of OH- / Mass of water (in kg)
Now, you can plug in the values into the equations and solve it step by step.