What is the pH of an aqueous solution of 0.10 M NaHC(3)H(2)O(4)?
Answer: 10.9
What I did:
NaHC(3)H(2)O(4) + H(2)O--><--
Na + H(2)C(3)H(3)O(4) + OH
H(2)C(3)H(3)O(4) is Malonic acid with a Ka value of 1.5*10^(-3)
initial-change-equilibrium (i.c.e) table results: (x^2)/(0.10-x)
x= 8.16*10^(-7)
[OH] = x
pOH = -log(x)
pH = 14-pOH
pH = 7.91
not sure what I did wrong.
I don't believe 10.9 is the correct answer.
If this is malonic acid, then NaHmal is the half neutralized salt of H2Mal and as such behaves as an amphiprotic salt. For simplified solutions the pH is given by 1/2(pK1 + pK2). For solutions that may not meet the criteria for that equation, one may use
(H+) = sqrt[(k2*C + Kw)/(1+ term)] where
term = C/k1
I worked the problem both ways using 2.83 for pk1 and 5.69 for pK2 and obtained pH = 4.26 for both equations. For what it's worth, your math is right but this is not the simple hydrolysis of a base HC3H2O4^- which is the basis of your solution.
After thinking about this for 2-4 hours, are you sure this is malonic acid? Malonic acid is H2C3H2O4 and the salts would be NaHC3H2O4 and Na2C3H2O4.
I guess not; I thought it was, but I could be wrong. If it's not malonic what do you think it is?
To determine the pH of the aqueous solution, you correctly set up an initial-change-equilibrium (i.c.e) table using the reaction:
NaHC3H2O4 + H2O ⇌ Na+ + HC3H3O4- + OH-
The dissociation of the NaHC3H2O4 produces HC3H3O4- and OH-. Since HC3H3O4- is a weak acid, you can use its Ka value to calculate the concentration of OH- and ultimately find the pH of the solution.
However, there seems to be a minor mistake in your calculations. Let's go through the correct process step by step:
1. Write the balanced chemical equation:
NaHC3H2O4 + H2O ⇌ Na+ + HC3H3O4- + OH-
2. Set up the initial concentration values in the i.c.e. table:
NaHC3H2O4: 0.10 M
H2O: Not included since it is a solvent and does not affect the equilibrium.
Na+: 0 M (assumed negligible)
HC3H3O4-: 0 M (initially, no dissociation)
OH-: 0 M (initially, no dissociation)
3. Determine the change in concentration (x):
NaHC3H2O4: -x (since it dissociates into HC3H3O4-)
H2O: No change (since it is a solvent)
Na+: +x (formed as a product)
HC3H3O4-: +x (formed as a product)
OH-: +x (formed as a product)
4. Set up the equilibrium expression using the Ka value:
Ka = ([HC3H3O4-][OH-])/[NaHC3H2O4]
Plugging in the initial concentrations and changes:
1.5 x 10^-3 = (x * x) / (0.10 - x)
5. Solve for x:
Since the value of x is small compared to 0.10 (which is the initial concentration of NaHC3H2O4), you can approximate 0.10 - x as 0.10:
1.5 x 10^-3 = (x * x) / (0.10)
x^2 = 1.5 x 10^-4
x = √(1.5 x 10^-4) ≈ 3.87 x 10^-3
Note: Make sure to take the square root of 1.5 x 10^-4, not 1.5 x 10^-3.
6. Calculate pOH:
pOH = -log(x) = -log(3.87 x 10^-3) ≈ 2.41
7. Calculate pH:
pH = 14 - pOH = 14 - 2.41 ≈ 11.59
So, the pH of the aqueous solution of 0.10 M NaHC3H2O4 is approximately 11.59, not 10.9.