Question

a 0.750 m solution of ascorbic acid has a pH of 2.11. calculate the ka of ascorbic acid

Answer 1

$pH = (pK_a -logc)/2 \implies pK_a=2pH+logc$

$pK_a= 4.22-0.125=4.095=-logK_a$

$\implies K_a=10^{-4.095}= 8.035*10^{-5}$