What is the pH of a solution containing 0.81M HA and 0.35M A- ? Ka HA=5.45x10-8.
This is a buffer solution.
(A^-) is 0.35 M in the solution. (HA) is 0.81 M
.................HA ==> H^+ + A^-
I.................0.81......0.........0.35
C................-x...........x..........x
E..............0.81-x.......x........0.35+x
Ka = (H^+)(A^-)/(HA)
Plug the E line nto the Ka expression and solve for (H^+), then convert it to pH.
You may also use the Henderson-Hasselbalch equation of
pH = pKa + log (base)/(acid) where base is the A^- and the acid is HA
Why did the acid go to the comedy club? Because it had a pH-unny joke to share! Now, let's figure out the pH of this solution.
First, let's find the concentration of H+ ions produced by HA. We can use the equation Ka = [H+][A-]/[HA]. Since [A-] is given as 0.35 M and [HA] is given as 0.81 M, we'll solve for [H+].
Ka = [H+][A-]/[HA]
5.45x10^-8 = [H+](0.35)/(0.81)
Now, let's solve for [H+]:
[H+] = (5.45x10^-8)(0.81)/(0.35)
[H+] ≈ 1.26x10^-7 M
Since pH = -log[H+], we can calculate the pH of the solution:
pH = -log(1.26x10^-7)
pH ≈ 6.90
So, the pH of the solution is approximately 6.90.
To find the pH of a solution containing a weak acid and its conjugate base, we will use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
Ka for HA = 5.45x10^-8
[A-] = 0.35 M
[HA] = 0.81 M
Step 1: Calculate pKa
pKa = -log(Ka)
pKa = -log(5.45x10^-8)
pKa ≈ 7.263
Step 2: Calculate [A-]/[HA]
[A-]/[HA] = 0.35 M / 0.81 M
[A-]/[HA] ≈ 0.4321
Step 3: Calculate pH using Henderson-Hasselbalch equation
pH = 7.263 + log(0.4321)
pH ≈ 7.263 + (-0.3635)
pH ≈ 6.8995
Therefore, the pH of the solution is approximately 6.8995.
To find the pH of the solution, we need to first determine the concentration of A- and HA in the solution. This can be done by using the equation for the dissociation of the acid HA:
HA ⇌ H+ + A-
The dissociation constant (Ka) for the acid HA is given as 5.45x10-8. It can be used to calculate the concentrations of H+ and A- in the solution.
Let's assume that x is the concentration of H+ (or A-) in moles per liter. This means that the concentration of A- is 0.35 M + x M, since some of the initial concentration of HA will dissociate to form A-.
Using the dissociation equation and the Ka expression, we can write the expression for the equilibrium constant (Keq) of the reaction:
Ka = [H+][A-]/[HA]
Substituting the values, we get:
5.45x10-8 = x * (0.35 + x) / 0.81
Simplifying the equation, we have:
5.45x10-8 = (0.35x + x^2) / 0.81
Rearranging the equation, we get a quadratic equation:
0 = x^2 + 0.35x - 5.45x10-8 * 0.81
Solving this quadratic equation will give us the value of x, which represents the concentration of H+ (or A-). Once we have the H+ concentration, we can calculate the pH using the equation:
pH = -log[H+]
Note: It's important to consider the assumption made while solving the quadratic equation. In this case, since the initial concentration of HA is much larger than the value of Ka, we can assume that x is small and can be ignored when added to 0.35 M.
By solving the quadratic equation and calculating the concentration of H+, we can find the pH of the solution.