The solubility of Phenobarbital (weak acid,MW=232 g/mole) in water at 25°C is 0.

Question

   The solubility of Phenobarbital (weak acid,MW=232 g/mole) in water at 25°C is 0.14% (w/v). What isthe pH of the saturated solution? Ka=3.9x10-8 M. The answer is 4.81, please explain what equation to use tosolve the problem and how to work the problem, thanks    The solubility of Phenobarbital (weak acid,MW=232 g/mole) in water at 25°C is 0.14% (w/v). What isthe pH of the saturated solution? Ka=3.9x10-8 M. The answer is 4.81, please explain what equation to use tosolve the problem and how to work the problem, thanks

Explanation / Answer

First assume that you have 100 mL of solution. You would thenhave 0.14 g of solute in 100 mL of a saturated solution. Convert this concentration to molarity: (0.14 g/100 mL)(1 mol/232 g)(1000 mL/1 L) = 0.00603 M This is the concentration of Phenobarbital in solution. Nowuse the definition of Ka to find the concentration ofH3O+. Ka = ([H3O+][conjugate base])/[acid] Since the Ka is such a small number, you can assume that theconcentration of Phenobarbital at equilibrium is the same as theinitial concentration. The equation simplifies to: Ka = x2/[acid] Plug in your values and solve for x: x = 1.53*10-5 This value is the concentration of H3O+ insolution. To get the pH just take the negative log of thisvalue: pH = -log(1.53*10-5) = 4.81 Hope this helps.

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