The pH of a 0.175 M aqueous solution of a weak acid is 3.52. What is Ka for this acid ?
Ka=[H+][attached-]/(.175-x)
or
ka= x^2/(.175-x)
but pH=-logx, or x=10^-3.52
so x^2=(3.04E-4)^2=about 9E-8 check that
Ka= (9E-8)/(.175-.000304)
ka=9E-8/.175 =1.73E-3
check all this, I did most of it in my head.
thank you so much!
To find the Ka (acid dissociation constant) of a weak acid given its pH and concentration, you can use the Henderson-Hasselbalch equation. This equation relates the pH, pKa, and the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is as follows:
pH = pKa + log ([A-]/[HA])
Where:
pH = the given pH of the solution
pKa = the negative logarithm of the acid dissociation constant
[A-] = concentration of the conjugate base
[HA] = concentration of the acid
In this case, you are given the pH of the solution (3.52) and the concentration of the acid (0.175 M).
However, in order to proceed, we need to know the ratio of the conjugate base to the acid concentration ([A-]/[HA]). This information is not provided in the question, so we will have to assume that the initial concentration of the conjugate base ([A-]) is negligible compared to the concentration of the acid ([HA]).
Therefore, we can simplify the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
pH = pKa + log ([A-]/0.175)
To solve for pKa, we rearrange the equation:
pKa = pH - log ([A-]/0.175)
Now that we have the values for pH and [HA] (0.175 M), we can substitute them into the equation:
pKa = 3.52 - log ([A-]/0.175)
Thus, to find Ka, we need to calculate the antilog of pKa:
Ka = 10^(-pKa)
Substitute the value of calculated pKa into this equation and solve. Remember to use logarithmic functions in your calculations.