Calculate the pH and molar concentration of each protonated form for a 0.133 M s

Question

Calculate the pH and molar concentration of each protonated form for a 0.133 M solution of NaHA.
pH:
H2A:
HA-:
A2-:
Calculate the pH and molar concentration of each protonated form for a 0.133 M solution of Na2A.
pH:
H2A:
HA-:
A2-:

Explanation / Answer

For H2A it follows the usual form of ionizsation as if you had a monoprotic acid. ..........H2A ==> H^+ + HA^- I.......0.133M....0......0 C.........-x......x......x E........0.133-x..x.......x Ka1 = (H^+)(HA^-)/(H2A) Substitute and solve for x = (H^+) = (HA^-) and convert to pH. (H2A) = 0.133-x. I strongly suspect you will need t use the quadratic equation to solve this because ka1 is relatively large. For NaHA. (H^+) = sqrt [(k2(HA^-)+Kw)/(1+(HA^-)/k1)] and convert to pH. For Na2A, set up an ICE chart for the hydrolysis of A^2-. I'm not getting the right answers for the pH, [H2A], and [HA^-] in part a (0.133 M solution for H2A) For H2A, I did 4.5*10^-2=x^2/(.133-x) Using the quadratic formula: x^2+.045x-.005985=0 x=.07147 [H^+]=[HA^-]=.07 M pH=-log(.072)=1.17 Finding [H2A]=.133 M-.07 M=.06 M Finding [A^2-]=Ka2*[HA^-]/[H^]=5.9E-8 M

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