What is the pKa for the acid HA if a solution of 0.65 M HA and 0.85 M NaA has a pH of 4.75?
~~~I know you find the ph by doing -log of pka; how do you do the reverse when you have molarity? I am completely lost. :-(
You're trying to make this a hard problem when it isn't.
This is a buffered solution, use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid)
They give the pH, base is 0.85 and acid is 0.65. Substitute and solve for pKa.
pKa = 4.63
To determine the pKa for the acid HA, you can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this case, the acid HA is in equilibrium with its conjugate base A-. Assuming complete ionization of NaA, you can consider [A-] to be 0.85 M, and [HA] to be 0.65 M.
Given that the pH is 4.75, you can rearrange the Henderson-Hasselbalch equation to solve for pKa:
pKa = pH - log([A-]/[HA])
Substituting the given values, we have:
pKa = 4.75 - log(0.85/0.65)
Calculating the ratio [A-]/[HA]:
[A-]/[HA] = 0.85/0.65 = 1.31
Therefore, the pKa for the acid HA is:
pKa = 4.75 - log(1.31)
You can use a scientific calculator or a logarithm table to calculate the logarithm and subtract it from 4.75 to obtain the pKa value.
To find the pKa of the acid HA, we need to first understand the relationship between pH, pKa, and the concentration of the acid and its conjugate base.
The pKa value represents the acidity/basicity of an acid-base system. It is defined as the negative logarithm (base 10) of the acid dissociation constant (Ka). Mathematically, pKa = -log10(Ka).
In this case, you are given the pH of a solution containing 0.65 M HA and 0.85 M NaA. To determine the pKa of HA, you need to use the Henderson-Hasselbalch equation, which relates the pH, pKa, and the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the conjugate base (in this case, NaA), and [HA] represents the concentration of the acid (0.65 M HA).
Since you have the pH value (4.75) and the concentrations of HA and NaA, you can rearrange the equation to solve for pKa:
pKa = pH - log([A-]/[HA])
Now, substitute the given values into the equation:
pKa = 4.75 - log(0.85/0.65)
Using a calculator or log table, calculate the ratio of [A-]/[HA]: 0.85/0.65 ≈ 1.31
Substitute this value into the equation:
pKa = 4.75 - log(1.31)
Finally, calculate the logarithm: log(1.31) ≈ 0.117
pKa = 4.75 - 0.117
The result is the pKa value for the HA acid in the given solution.