Given a diprotic acid, H2A, with two ionization constants of Ka1=3.9E-2 and Ka2=5.9E-8.
Calculate the pH and molar concentration of each protonated form for a 0.135 M solution of:
(a) H2A
(b) NaHA
(c) Na2A
To calculate the pH and molar concentration of each protonated form for a 0.135 M solution of each compound, we'll use the equations relating acid dissociation constants and pH.
Let's start with (a) H2A:
(a) H2A:
H2A <--> H+ + HA-
First, we'll calculate the concentration of each protonated form using the equation:
[H+] = [HA-] = sqrt(Ka1 * [H2A])
where [H2A] is the initial concentration of H2A.
[H+] = [HA-] = sqrt((3.9E-2) * (0.135))
[H+] = [HA-] = sqrt(0.005265)
[H+] = [HA-] = 0.0725 M
To calculate the pH, we'll use the equation:
pH = -log[H+]
pH = -log(0.0725)
pH ≈ 1.14
So, for a 0.135 M solution of H2A, the pH is approximately 1.14 and the molar concentration of each protonated form [H+] and [HA-] is approximately 0.0725 M.
Now, let's move on to (b) NaHA:
(b) NaHA:
NaHA <--> Na+ + HA-
Since NaHA is a salt, it will completely dissociate in water, meaning the concentration of Na+ will be equal to the concentration of HA-. So, the concentration of HA- will be 0.135 M.
To calculate the pH, we'll use the equation:
pH = -log([H+])
Since HA- is the only protonated form, and its concentration is 0.135 M, this means [H+] is also 0.135 M.
pH = -log(0.135)
pH ≈ 0.87
Thus, for a 0.135 M solution of NaHA, the pH is approximately 0.87 and the molar concentration of HA- (also [H+]) is 0.135 M.
Finally, let's calculate for (c) Na2A:
(c) Na2A:
Na2A <--> 2Na+ + A2-
Similar to (b), Na2A is a salt, and 2Na+ will have the same concentration as A2-. So, the concentration of A2- will be 0.135 M.
To calculate the pH, we'll use the equation:
pH = -log([H+])
Since A2- is the only protonated form, and its concentration is 0.135 M, this means [H+] is also 0.135 M.
pH = -log(0.135)
pH ≈ 0.87
Therefore, for a 0.135 M solution of Na2A, the pH is approximately 0.87 and the molar concentration of A2- (also [H+]) is 0.135 M.
To calculate the pH and molar concentration of each protonated form, we need to consider the ionization constants of the diprotic acid and its salt solutions. Here's how you can find the answers for each case:
(a) H2A:
For a 0.135 M solution of H2A, we assume all the acid is in the H2A form at the beginning. Since H2A is a diprotic acid, it can form three different species: H2A (undissociated), HA^- (monoanion), and A^2- (dianion).
First, we need to calculate the concentration of H2A that dissociates to form HA^-. Using the ionization constant Ka1=3.9E-2, we can set up an equilibrium expression:
[H2A] -> [HA^-] + [H+]
Let x be the concentration of [HA^-], then [H2A] = 0.135 – x and [H+] = x.
The equilibrium constant expression can be written as:
Ka1 = [HA^-][H+] / [H2A]
Substituting the values:
3.9E-2 = x * x / (0.135 – x)
Since we know Ka1 is small compared to the initial concentration of the acid (0.135 M), we can assume that x is much smaller than 0.135. It allows us to simplify the equation as:
3.9E-2 = x^2 / 0.135
Solving this equation gives:
x ≈ 0.047
So, the concentration of HA^- is approximately 0.047 M, and the concentration of H2A is 0.135 – 0.047 = 0.088 M.
To calculate the pH, we need to find the concentration of H+ ions. At equilibrium, the concentration of H+ is equal to the concentration of HA^-. Therefore, the pH is given by:
pH = -log[H+] = -log(0.047) ≈ 1.33
(b) NaHA:
Now, let's consider a 0.135 M solution of NaHA. Since NaHA is the salt formed by the reaction of the diprotic acid H2A with NaOH, it dissociates completely into its ions in water. The Na+ ion is a spectator ion and does not affect pH.
NaHA -> Na+ + HA^-
Since NaHA dissociates completely, the concentration of HA^- is 0.135 M and the concentration of Na+ is also 0.135 M.
To calculate the pH, we need to find the concentration of H+. Since the concentration of HA^- is the same as the concentration of H+ ions, the pH can be determined from:
pH = -log[H+] = -log(0.135) ≈ 0.87
(c) Na2A:
Lastly, let's consider a 0.135 M solution of Na2A. Similar to NaHA, Na2A dissociates completely into its ions in water.
Na2A -> 2Na+ + A^2-
The concentration of A^2- is equal to the concentration of Na2A, which is 0.135 M. The concentration of Na+ is twice the concentration of Na2A since there are two Na+ ions produced for each Na2A molecule in the salt.
To calculate the pH, we need to find the concentration of H+ ions. In this case, there is no acidic species (such as H2A or HA^-) to provide H+ ions. Therefore, the pH of a solution of Na2A cannot be determined from the given information.
For H2A it follows the usual form of ionizsation as if you had a monoprotic acid.
..........H2A ==> H^+ + HA^-
I.......0.135M....0......0
C.........-x......x......x
E........0.135-x..x.......x
Ka1 = (H^+)(HA^-)/(H2A)
Substitute and solve for x = (H^+) = (HA^-) and convert to pH. (H2A) = 0.135-x. I strongly suspect you will need t use the quadratic equation to solve this because ka1 is relatively large.
For NaHA.
(H^+) = sqrt [(k2(HA^-)+Kw)/(1+(HA^-)/k1)] and convert to pH.
For Na2A, set up an ICE chart for the hydrolysis of A^2-.
I'm not getting the right answers for the pH, [H2A], and [HA^-] in part a (0.135 M solution for H2A)
For H2A, I did
3.9E-2=x^2/(.135-x)
Using the quadratic formula:
x^2+.039x-.005265=0
x=.06725
[H^+]=[HA^-]=.06725 M
pH=-log(.06725)=1.17
Finding [H2A]=.135 M-.06725 M=.06775 M
Finding [A^2-]=Ka2*[HA^-]/[H^]=5.9E-8 M