How would you write the Henderson-Hasselbalch equation for a solution of propanoic acid (#CH_3CH#2CO#2H#, pKa = 4.874 ) using HA, A–, and the given pKa value in the expression?

1 Answer
Feb 4, 2017

Answer:

#pH=pK_a+log_10([[A^-]]/[[HA]])# is the general equation.........

Explanation:

#pH=4.874+log_10([[A^-]]/[[HA]])#

Of course, we don't know the concentrations of the starting propanoic acid. If there are equimolar quantities of propanoic acid and propanoate ion then:

#pH=pK_a=4.874#. Why so?