For the diprotic weak acid H2A Ka1 25SolutionFor H2A I did

For the diprotic weak acid H2A, Ka1 = 2.5

For H2A, I did
2.5*10^-5=x^2/(0.075-x)

Using the quadratic formula:
x^2+2.5*10^-5x-0.1875*10^-5=0
x=1.38*10^-3
[H^+]=[HA^-]=1.38*10^-3 M

pH=-log(1.38*10^-3)=2.86

Finding [H2A]=.075 M-1.38*10^-3 M=0.0736 M

Finding [A^2-]=Ka2*[HA^-]/[H^]=7.9*10^-7 M