Question

Tartaric acid (“H₂Tart”, M.W. = 176.13) is a diprotic acid that is important in food preparation. (See McMurray & Fay Appendix C for its Ka values.) A solution of tartaric acid is produced by dissolving 2.90 g of H₂Tart in 1.000 L of water. What are the equilibrium concentrations of H₂Tart, HTart^-, Tart^_2-, and H⁺ in this solution?

Answer 1

Answer- Given, mass of H₂Tart = 2.90 g , volume = 1.00 L

First we need to calculate the moles

We know,

Moles of H₂Tart = 2.90 / 176.13 g.mol^-1

                         = 0.0165 moles

So, [H₂Tart] = 0.0165 moles / 1.00 L = 0.0165 M

No first we need to calculate the concentration of give species

tartaric acid K1 = 9.2*10^-4 , K2 = 4.3*10^-5

H₂Tart is weak acid, so we need to put ICE chart

      H₂Tart + H₂O -------> HTart ^- + H₃O⁺

I        0.0165                  0            0

C        -x                      +x            +x

E      0.0165-x               +x           +x

Ka1 = [HTart ^-] [H₃O⁺] / [H₂Tart]

9.2*10^-4 = (x) * (x) / (0.0165-x)

We can neglect x in 0.0165-x , since Ka1 value is too low

9.2*10^-4 * 0.0165 = x²

x = 0.00389 M

[H₃O⁺] = x = 0.00389 M

So, [H₂Tart] = 0.0165-x

                     = 0.0165 -0.00389

                     = 0.0126 M

x = [HTart^-] = 0.00389 M

Now second dissociation-

     HTart^- + H₂O ------> Tart^2- + H₃O⁺

I    0.00389                 0            0

C      -x                     +x         +x

E 0.00389–x           +x           +x

Ka2 = [Tart^2-] [H₃O⁺] / [HTart^-]

4.3*10^-5 = (x) * (x) / (0.00389-x)

We can neglect x in 0.00389-x , since Ka2 value is too low

4.3*10^-5 * 0.00389 = x²

x = 0.000409M

[H₃O⁺] = x = 0.000409 M

x = [Tart^2-] = 0.000409 M

So total [H₃O⁺] = 0.00389M + *0.000409 M

                       = 0.00430 M