Question

Part If the pH at one half the first and second equivalence points of a dibasic acid is 4.80 and 7.34, respectively, what are the values for pka and pKaz? Separate your answers with a comma. IO AD O ? Submit Request Answer Part D Using the pKar and pka2 values from the previous question, calculate the values for Kay and Ka2 Separate your answers with a comma. I ADV * * O ? Submit Request Answer

Answer 1

One half the equivalence point is known as midpoint. At this point, the concentration of ionised salt = deionised acid.

Since, pH = pKa + log {ionised salt / deionised acid}

=> pH = pKa + log (1)

=> pH = pKa + 0 = pKa.

Second Midpoint a equivalence point First equivalence point Midpoint volume of base Added (ML)

Part C.

pKa₁ = pH at equivalence point 1 = 4.80

pKa₂ = pH at equivalence point 2 = 7.34

Part D.

Now, pKa = - log (Ka)

=> Ka = 10 ^-pKa

So, Ka₁ = 10 ^-4.80 = 1.58 × 10^-5

Ka₂ = 10 ^-7.34 = 4.57 × 10^-8