It is in the ratio of: 1 to 27
The LCM is 7182.
Ka has an "a" k does not
Muqaddar Ka Faisla was created in 1987.
Naukar Biwi Ka was created in 1983.
All you need to do to get Ka is to take the antilog of the pKa.
pKa = -log Ka and thus Ka = 10^-pKaKa = antilog pKaKa = 7.76x10^-6
The Ka of hydrogen iodide is approx. 1010. The pKa of hydrogen iodide is approx. - 9.
pKa = -log KapKa = -log 5.4x10^-10pKa = 9.27
It refers to the acidity of the fatty acid (which make up the oils). Every fatty acid is composed of a non-polar long chain of hydrocarbons (carbon and hydrogen) and a polar head made up of Carboxylic ACID. Every acid has something called pKa which determines the acidity of that acid.The bigger the pKa (e.g. 25), the weaker the acid.The smaller ther pKa (e.g. 2), the stronger the acid.
At half titration pH=pKa (you need the pH from the graph of your titration, y axis) ph = pKa + log (base/acid) 10^-pKa = Ka Kw=Ka*Kb Kb=Kw/Ka Ka = Kw/Kb
The KB value is listed below: pKa + pKb - 14 pKa = -log10 (Ka), Ka of acetic acid = 1.8*10^-5 pKa = 4.74, pkb = 9.255 9.255 = -log10(Kb), Kb = 5.56*10^-10
The pKa of sulfonic acid is < 0
pKa < 0.0 Ka > 1.0
The pKa of the fluorosulfuric acid is -10; HSO3F is a very strong acid, a so-called superacid. Ka is the dissociation constant; pKa is the decimal logarithm of Ka.
PKa = -log Ka so if you multiply across by -1 and then taking the antilog you can get Ka Ka.Kb = Kw where Kw = 1.0 x 10^14 PKa + PKb = PKw = 14 that should give you a start.
Its an equation you can use to find the pH of a solution. it is.... --- pH = pKa + log (Base/Acid) --- these may help too Ka = 10^-pKa Kw = Ka*Kb