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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.
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
According to CRC reference data, the pKa of sulfamic acid is 1.05, giving a Ka of 11.2. This is a strong acid.
Ka = [H+][A-] / [HA] Hence [H+] = Ka[HA] / [A-] Remember pH = -log(10)[H+] 'logging' both sides. -log(10)[H+] = - log(10)Ka[HA] / [A-] By algebraic manipulation of log. pH = -log[A-]^-1 - logKa - log[HA] pH = log[A-] - logKa - log[HA] pH = pKa - log[HA]/[A-]
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
pKa = -log Ka and thus Ka = 10^-pKaKa = antilog pKaKa = 7.76x10^-6
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.
pKa = -log KapKa = -log 5.4x10^-10pKa = 9.27
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.
No. An acid with a large Ka is stronger. A lower pKa indicates a stronger acid.
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
As with any acid, that depends entirely on its concentration. The greater the concentration (until saturation) the lower the pH. Lactic acid is a "middle of the road" weak acid; all in equal concentrations, lactic acid is stronger than citric, weaker than acetic, and about the same as formic.
The pKa of sulfonic acid is < 0
According to CRC reference data, the pKa of sulfamic acid is 1.05, giving a Ka of 11.2. This is a strong acid.
Ka = [H+][A-] / [HA] Hence [H+] = Ka[HA] / [A-] Remember pH = -log(10)[H+] 'logging' both sides. -log(10)[H+] = - log(10)Ka[HA] / [A-] By algebraic manipulation of log. pH = -log[A-]^-1 - logKa - log[HA] pH = log[A-] - logKa - log[HA] pH = pKa - log[HA]/[A-]
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
Yes this is true. This is because for a buffer solution, Ka = [H+] ( [acid] ) / [salt] ) As such by mathematical manipulation, [H+] = Ka ( [salt]/ [acid] ) We must keep in mind that a good buffer must have equal concentrations of acid and salt so as to be able to resist pH change in both directions, by absorbing protons and hydroxide ions. As such the value of ( [salt] / [acid] ) will be 1 and can be cancelled from the equation. [H+] = Ka And, pH = pKa :)