Question

I was given this information and I need to draw a calibration curve of dissociation for carbon- Please help. I am having a difficult time getting started.

Find pH of 1.) 0.300 M H2CO3

2.) 0.300 M HCO3

3.) 0.300 M CO3^2

Answer 1

1. You need to know the K_a for H₂CO₃ to answer this question. This could be given to you in the question, or you have to look it up.
   
2. 
   
   
   H₂CO₃ is NOT a strong acid (so it will not dissociate completely). H₂CO₃ will only dissociate slightly into H⁺ and HCO₃^-.
   
   Because this is a weak acid, the formula for your equation is:
   
   Ka = [H⁺][HCO₃^-] / [H₂CO₃]
   
   Because the H⁺ and HCO₃^- ions are being formed in equal concentrations, call the concentration of each of these molecules "x".
   The initial concentration of H₂CO₃ is .1. After the reaction, it will lose x amount of substance, which will form the H+ and HCO3-.
   
   
   Here is your equation (with variables and numbers substituted in):
   Ka = 4.3 x 10^-7 = x²/(0.3 - x)
   
   Estimate that x is so small that 0.3 - x is around x. This is called the 5% rule, which you must check after the problem.
   
   Now, solve the equation for x.
   
   (4.3 x 10^-7 )(0.3) = x²
   1.29 x 10^-7 = x²
   x = 3.6 x 10^-4 = the [H⁺] concentration
   
   pH = -log[H+] = - log(3.6 x 10^-4 ) = 3.4

1. the ion does a hydrolysis
   HCO₃^- + H2O --> H₂CO₃ + OH^-
   
   Kh = Kwater / K_H2CO3 = 1 x 10^-14 / 4.4 × 10^–7 = 2.27 x 10^–8
   
   Kh = [ H₂CO₃] [OH^-] / [HCO₃^-]
   
   2.27 x 10^–8 = [ X] [X] / [0.30]
   
   X² = 6.81 x 10^–9
   
   X = [OH^-] = 8.25 x 10^–5
   
   pOH = -log (OH^-) = 4.08
   
   pH = 14 - pOH = 14 – 4.08 = 9.92

1. CO₃^2- + H₂O --> HCO₃^- + OH^-
   This is a salt hydrolysis problem; say 0.3 M of Na2CO3 was dropped in solution; since this is soluble, there would be 0.3 M of CO₃^2- in solution (refer to solubility rules). Since this salt is composed of the conjugates of a strong base (NaOH) and a weak acid (HCO₃^-), the solution will be basic. The assigning of Na+ was arbitrary, it would still be a hydrolysis reaction. Ka for HCO₃^- is 5.6e-11 at 25 degrees C, where e represents scientific notation.

    Ka Kb = Kw,

so Kb = Kw/Ka= (1e-14)/(5.6e-11) = 1.786e-4.

Since carbonate is the conjugate of a weak acid, it will be a relatively strong base, according to the hydrolysis equation (produces OH- ions, and due to the inverse proportionality of Ka and Kb). Anyway,

x²/(0.3 - x) = 1.786e-4

x = 8.875e-3 = [OH-]

pH = 14 - pOH = 14 + log(x) = 11.95.