Can someone help me with this question? This could help me with how to do another question, similar to this, for my homework.
1.Estimate the pH of
(a) 0.15 M NAHSO3 (aq)
(b) 0.050 M NaHSO3 (aq)
Thank you!
Of course, I can help you with that! To estimate the pH of a solution, you need to consider the dissociation of the given compound in water. In this case, we have sodium bisulfite, NaHSO3, which is an acidic salt.
To estimate the pH, we need to understand the dissociation of NaHSO3 in water. NaHSO3 ionizes in water to produce Na+ and HSO3-, which can further dissociate to produce H+ and SO3^2- ions. The H+ ions determine the acidity of the solution, so we need to focus on that.
Now, let's calculate the pH for each concentration:
(a) 0.15 M NaHSO3 (aq):
Since NaHSO3 is a salt, it undergoes partial hydrolysis in water. It reacts with water to produce H+ ions and bisulfite ions (HSO3-). Since the concentration is 0.15 M, we can assume it will partially dissociate.
NaHSO3 + H2O ⇌ HSO3- + Na+ + H+
The extent of dissociation depends on the acid dissociation constant (Ka) of bisulfite, which is 1.1 x 10^-7.
The pH can be estimated by calculating the concentration of H+ ions in the solution. We can assume that the concentration of HSO3- (bisulfite) at equilibrium will be the same as the initial concentration of NaHSO3. Therefore, [HSO3-] ≈ 0.15 M.
To calculate the concentration of H+ ions:
[H+] = K_a * [HSO3-]
[H+] ≈ (1.1 x 10^-7) * (0.15)
[H+] ≈ 1.65 x 10^-8 M
Since the concentration of H+ is much smaller than the concentration of NaHSO3, we can assume that the solution is only slightly acidic. Hence, the pH is estimated to be higher than 7, but still close to neutral.
(b) 0.050 M NaHSO3 (aq):
Following the same process as above, the concentration of HSO3- in this case will be approximately 0.050 M.
[H+] ≈ (1.1 x 10^-7) * (0.050)
[H+] ≈ 5.5 x 10^-9 M
Again, since the concentration of H+ ions is much smaller than the concentration of NaHSO3, we can estimate the pH to be slightly acidic.
Remember, this method provides an approximate estimation of the pH. For a more accurate calculation, you would need to consider the activity coefficients and ionic strength of the solution.
I'm not quite sure of how much "estimation" the author wants but this is what I would do.
.......HSO4 ==> H^+ + SO4^2-
I......0.15.....0......0
C.......-x......x......x
E....0.15-x.....x......x
K2 for H2SO4 = (x)(x)/(0.15-x)
I would solve that exactly which will give a quadratic equation. Estimate the answer may mean to assume 0.15-x = 0.15
The difference in the two answers for (H^+) is about 15% so the term estimate may very well mean to assume 0.15-x = 0.15 to avoid the quadratic equation.
Then pH = -log(H^+).