[johnnyjohn993123]

When 2.0x10^-2 mole of nircotic acid (monoprotic) is dissolve in 350mL of water the pH is 3.05. What is the Ka of nircotic acid??

My attempt:
Ka=α² * initial conc

Do I need to do this 2.0×10^-2 / 0.350 L? To get the conc? Isnt that there should be like they said liters of soln to get the Conc of the acid pleaase do help.

[mjc123]

Yes, that gives you the concentration.
You do not need to work out α. pH gives you [H⁺], and you can work out [A^-] and [HA] directly.

[johnnyjohn993123]

Yes, that gives you the concentration.
You do not need to work out α. pH gives you [H⁺], and you can work out [A^-] and [HA] directly.

I thought Molar concentration is said to be moles of solute/ Liters of solution ?
that is Liters of solution would contain two or more substance in it? 
That being said, wouldn't dividing moles of the substance by water (which is just a solvent) would generate somewhat the molality (i.e kg of solute/ L in solution)

[mjc123]

If the solution is dilute, it is reasonable to assume that the volume of the solution is the same as the volume of water; so 0.02 mol solute in 350 mL water gives 350 mL solution. (At any rate, you have no information to calculate the solution volume more accurately.)
Molality is moles of solute per kg of solvent. For a dilute aqueous solution, this is to a first approximation numerically equal to the molarity, as we assume 1 kg water gives 1 L solution.

[johnnyjohn993123]

If the solution is dilute, it is reasonable to assume that the volume of the solution is the same as the volume of water; so 0.02 mol solute in 350 mL water gives 350 mL solution. (At any rate, you have no information to calculate the solution volume more accurately.)
Molality is moles of solute per kg of solvent. For a dilute aqueous solution, this is to a first approximation numerically equal to the molarity, as we assume 1 kg water gives 1 L solution.

what if 0.05 mol of AgNO₃ is dissolved in 500mL of 3.0M NH₃ . like this complex formation do you do an ice table in moles and divide the desired substance (in mole) with 0.5L to get the desired concentration at equilibrium of the substance?

[mjc123]

I would do the ice table in concentrations throughout.