A solution that contains 2.047 x 10^-1 M of acid, HA, and 1.808 x 10^-1 M of its conjugate base, A-, has a pH of 4.196. What is the pH after 1.629 x 10^-3 mol NaOH is added to 5.961 x 10^-1 L of this solution?
(HA) = 0.2047
(A^-) = 0.1808
moles HA = 0.2047M*0.5961L = 0.12202
moles A^- = 0.1808 x 0.5961 = 0.10777
pH = pKa + log (base)/(acid)
The problem states the pH as well as the beginning HA and A. Substitute those and solve for pKa. Then look at the ICE chart below.
............HA + OH^- ==> A^- + H2O
initial.0.12202...0...0.10777....
add............0.001629............
change..-.001629.-.001629..+.001629.....
equil......?......0.......?
Fill in the ? spots and substitute into the Henderson-Hasselbalch equation to solve for new pH. Use pKa from the first calculation above. Post your work if you need additional help. I have carried more digits than allowed; you should round as needed.
What would the Ka for this be?
Oops, sorry. I understand now, thank you.
I calculated pKa = 4.25 so Ka = 5.624E-5
To find the pH after NaOH is added to the solution, we need to consider the reaction that occurs between the acid (HA) and the base (NaOH). The reaction can be represented by the following balanced equation:
HA + NaOH -> H2O + NaA
First, let's determine the initial number of moles of acid (HA) and its conjugate base (A-) in the solution:
Number of moles of HA = concentration of HA x volume of solution
= (2.047 x 10^-1 M) x (5.961 x 10^-1 L)
= 1.221 x 10^-1 mol
Number of moles of A- = concentration of A- x volume of solution
= (1.808 x 10^-1 M) x (5.961 x 10^-1 L)
= 1.080 x 10^-1 mol
Since NaOH is a strong base, it will react completely with the acid (HA). Therefore, the number of moles of HA will decrease by the amount of NaOH added, and the number of moles of A- will increase by the same amount.
Now, let's calculate the final number of moles of HA and A- after the addition of NaOH:
Final number of moles of HA = initial moles of HA - moles of NaOH
= (1.221 x 10^-1 mol) - (1.629 x 10^-3 mol)
= 1.205 x 10^-1 mol
Final number of moles of A- = initial moles of A- + moles of NaOH
= (1.080 x 10^-1 mol) + (1.629 x 10^-3 mol)
= 1.096 x 10^-1 mol
Since we now have the final moles of HA and A-, we can calculate the concentration of each species in the solution after the addition of NaOH:
Concentration of HA = moles of HA / volume of solution
= (1.205 x 10^-1 mol) / (5.961 x 10^-1 L)
= 2.022 x 10^-1 M
Concentration of A- = moles of A- / volume of solution
= (1.096 x 10^-1 mol) / (5.961 x 10^-1 L)
= 1.838 x 10^-1 M
Now, to calculate the pH of the solution after the addition of NaOH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Given that the pH is 4.196 and the concentration of A- and HA have been calculated, we can solve for pKa:
4.196 = pKa + log ((1.838 x 10^-1) / (2.022 x 10^-1))
Rearranging the equation to solve for pKa:
pKa = 4.196 - log ((1.838 x 10^-1) / (2.022 x 10^-1))
Using the given pH and the concentrations of A- and HA, we can now calculate the pKa.