[bassman]

This is my second attempt to get help , I read the info that the other person suggested but some how it's not sinking in and I think I'm making it to complicated. If someone would work out a couple of problems for me maybe I con understand. I promise it's not homework as this id independent study so I can get into the class. Any way calculate the ph of 5.0g of HBr in a 100 ml aqueous solution,and 2.0 g of NaOH- in a 50 ml aqueous solution. Please if you can show work. 2nd , are these acid ,basic ,or neutral? A).10 M HCl( I say ,acid) B) .20 m NaOH (I say,base) C) 6.8 x 10-7 m H3O+ (I say, Acid) D) H+ = .00056 m  (I say acid) I really need help I've worked out of 3 o-chem books and I'm lost .Thanx ,Robert

[mike]

HBr ----> H⁺ + Br^-

molecular weight of HBr = 80.9 g/mol (calculated from periodic table)
mass of HBr = 5.0g (given in question)

number of moles of HBr; n = m/M = 5.0/80.9 = 0.062 moles

volume of solution = 100mL (given in question)

concentration of HBr; c = n/v = 0.062/0.100 = 0.62 mol/L

pH = -log₁₀[H⁺]

concentration of HBr = [HBr] = [H⁺] = 0.62mol/L

therefore:

pH = -log₁₀(0.62)

pH = 0.2

[wereworm73]

As for the NaOH solution, you can get the concentration of OH^-, convert it to pOH and then determine the pH with the formula pH + pOH=14.



And you're right about those being acids & bases.