question archive I would not typically teach this to my high school students, so I looked around and found a great explanation on you tube
I would not typically teach this to my high school students, so I looked around and found a great explanation on you tube
Subject:ChemistryPrice: Bought3
I would not typically teach this to my high school students, so I looked around and found a great explanation on you tube. Since, in a polyprotic acid the first hydrogen will dissociate faster than the others, If the Ka values differ by a factor of 10 to the third power or more, it is possible to approximately calculate the by using only the Ka of the first hydrogen ion. For example:
Pretend that ##H_2X## is a diprotic acid. Look up on a table the Ka1 for the acid.
If you know the concentration of the of the acid, say it is 0.0027M and the ##Ka_1## is ##5.0 x 10^(-7)##. Then you can set up your equation as follows; ##H_2X## --> ##H^(+1)## + ##HX^(-1)## with ##Ka_1## = ##5.0x10^(-7)## Using the formula: Ka=(products) / (reactants):
##5.0x10^(-7)## = (##x^2##)/(0.0027) Then solve for x and you have your hydrogen ion concentration. Since pH = -log of hydrogen ion concentration, you can now calculate the pH.
