Calculate the [H3O+] and pH of each polyprotic acid solution.
(b) 0.116 M H3C6H5O7
I did part (a) just fine, but for some reason can't figure (b) out. Can someone help me with just the [H3O+] part to get me started?
I already got 9.3e-3 and 1.4e-3 without success and only have one try left.
The Ka values according to the appendix given are Ka1=7.4x10-4, Ka2=1.7x10-5, Ka3=4.0x10-7.
Thank you so much.
The [H3O+] of 0.116 M H3C6H5O7 can be calculated using the Henderson-Hasselbalch equation:
[H3O+] = Ka1 x [H3C6H5O7] / (Ka1 - Ka2)
= 7.4 x 10-4 x 0.116 / (7.4 x 10-4 - 1.7 x 10-5)
= 0.0114 M
To calculate the [H3O+] of the solution, you can set up an ICE (Initial, Change, Equilibrium) table and use the Ka values provided.
The given polyprotic acid is H3C6H5O7, which has three dissociation steps:
H3C6H5O7 ⇌ H+ + HC6H5O7- ... Ka1
HC6H5O7- ⇌ H+ + C6H5O7^2- ... Ka2
C6H5O7^2- ⇌ H+ + C6H5O7^3- ... Ka3
Let's focus on the first step dissociation:
H3C6H5O7 ⇌ H+ + HC6H5O7-
Using the Ka value given (Ka1 = 7.4x10^-4), we can set up the ICE table as follows:
Initial: 0.116 M 0 M 0 M
Change: -x +x +x
Equilibrium: 0.116-x x x
Now we can write an expression for Ka1:
Ka1 = [H+][HC6H5O7-] / [H3C6H5O7]
Since the concentration of HC6H5O7- at equilibrium is equal to [H+], we can simplify the expression to:
Ka1 = [H+]^2 / (0.116 - x)
Since we assume that x is very small compared to the initial concentration (0.116 M), we can approximate 0.116 - x to be approximately 0.116.
Now, solve for x by using the given Ka1 value and the equation:
7.4x10^-4 = x^2 / 0.116
Rearrange the equation to solve for x:
x^2 = 7.4x10^-4 * 0.116
x^2 = 8.584x10^-5
x ≈ 9.27x10^-3
Therefore, the approximate [H3O+] concentration is 9.27x10^-3 M.
Note: The pH can be calculated from the [H3O+] using the formula -log[H3O+].
To calculate the [H3O+] of a polyprotic acid solution, you need to consider all the dissociation steps and their respective equilibrium constants (Ka values).
The polyprotic acid in question is H3C6H5O7 (citric acid) with three dissociation steps indicated by the three Ka values given: Ka1, Ka2, and Ka3.
The dissociation reactions for each step can be written as follows:
Step 1: H3C6H5O7 ⇌ H+ + H2C6H5O7-
Ka1 = [H+][H2C6H5O7-]/[H3C6H5O7]
Step 2: H2C6H5O7- ⇌ H+ + HC6H5O7^2-
Ka2 = [H+][HC6H5O7^2-]/[H2C6H5O7-]
Step 3: HC6H5O7^2- ⇌ H+ + C6H5O7^3-
Ka3 = [H+][C6H5O7^3-]/[HC6H5O7^2-]
To calculate [H3O+] for the given concentration of 0.116 M H3C6H5O7, you need to consider the equilibrium expressions and concentrations of the species involved.
Let x represent the concentration of [H+] or [H3O+] at equilibrium for all three steps since they are all related.
For Step 1, since H3C6H5O7 only has one dissociable proton, the concentration of [H3C6H5O7] is equal to the initial concentration, which is 0.116 M. Therefore, [H3O+] = x and [H2C6H5O7-] = x.
For Step 2, the concentration of [H2C6H5O7-] at equilibrium is equal to the amount dissociated in Step 1 (x), so [H2C6H5O7-] = 0.116 M - x.
For Step 3, the concentration of [HC6H5O7^2-] at equilibrium is equal to the amount dissociated in Step 2 (x), so [HC6H5O7^2-] = 0.116 M - x.
Since each Ka value is small compared to the initial concentration of H3C6H5O7, you can make the approximation that x is small compared to 0.116 M.
Using this approximation, you can simplify the calculations. Substitute the relevant values into the equilibrium expressions and solve for x.
For Step 1:
7.4x10^(-4) = x^2 / (0.116 - x)
For Step 2:
1.7x10^(-5) = x^2 / (0.116 - x)
For Step 3:
4.0x10^(-7) = x^2 / (0.116 - x)
You can solve each equation separately to obtain the value of x. Remember to check and see if the approximation x << 0.116 M holds true after solving. If it does, you can proceed with using the value of x as [H3O+].
Once you have found the value of [H3O+], you can calculate the pH using the formula pH = -log([H3O+]).
Note: The calculations provided are based on the simplifying approximation and may not be completely accurate. Please double-check the calculations and assumptions before finalizing your answer.