Acid in vinegar has pka value of 1.85 x 10^-5 at 25oC find the pkb value of its conjugate base?
=4.6POH PH=9.4
To find the pKb value of the conjugate base of an acid, we can use the relationship between pKa and pKb.
pKw = pKa + pKb
Since pKw (the ionization constant of water) is a constant value of 14 at 25°C, we can rearrange the equation to solve for pKb:
pKb = pKw - pKa
First, we convert the given pKa value to Ka using the equation:
Ka = 10^(-pKa)
Substituting the given pKa value:
Ka = 10^(-1.85 x 10^-5)
Next, we calculate Kw using the equation:
Kw = Ka * Kb
Since Kw is a constant value of 1.0 x 10^-14 at 25°C, we can rearrange the equation to solve for Kb:
Kb = Kw / Ka
Substituting the calculated Ka value:
Kb = (1.0 x 10^-14) / (10^(-1.85 x 10^-5))
Finally, we convert Kb to pKb using the equation:
pKb = -log10(Kb)
Calculating pKb using the calculated Kb value will give us the answer to our question.