what is the pH of a solution that is .15 M in HOCl and .25 M NaOCl after .05 mol HCl/L has been bubbled into the solution?
this is what I did:
HOCl + H2O ----> H3O+ + OCl-
.15 .25
-.05 +.05
---------------------------
.1 .30
3.5E-8=[H3O][.3]/[.1]
H3O= 1.167E-8
pH= -log(1.67E-8)
pH=7.78
But the answer was 7.46. What did I do wrong?
When you add 0.05M H^+, that decreases NaOCl by 0.05 (to make it 0.20M), and you increase HOCl by 0.05 (to make it 0.20M) so you have
3.5E-8=(H^+)(0.20)/(0.20)
3.5E-8 = (H^+) and pH = 7.46.
Do you use the Henderson-Hasselbalch equation. That makes buffer problems a lot faster to work.
pH = pKa + log[(base)/(acid)]
pH = 7.46 + log (0.20/0.20)
pH = 7.46.
To calculate the pH of the solution after adding HCl, you need to consider the fact that HCl is a strong acid that completely dissociates in water.
The balanced equation for the reaction between HCl and HOCl is:
HCl + HOCl → H2O + ClO-
First, let's calculate the change in concentrations of HOCl and NaOCl after adding HCl:
Initial concentration of HOCl = 0.15 M
Change in concentration of HOCl = -0.05 M (from 0.15 M to 0.1 M)
Initial concentration of NaOCl = 0.25 M
Change in concentration of NaOCl = +0.05 M (from 0.25 M to 0.3 M)
Now, let's calculate the concentration of [H3O+] using the HCl added:
[HCl] = 0.05 M (since 0.05 mol HCl is added to the solution)
[H3O+] = [HCl] = 0.05 M
Next, let's calculate the concentration of [OCl-]:
[OCl-] = Change in concentration of NaOCl = 0.05 M
Now, let's use the equilibrium expression for the dissociation of water to calculate [H3O+]:
[H3O+] * [OCl-] / [HOCl] = 3.5E-8
Substituting the values we calculated:
(0.05 M) * (0.05 M) / (0.1 M) = 3.5E-8
[H3O+] = 1.75E-8 M
Finally, calculate the pH using the equation pH = -log[H3O+]:
pH = -log(1.75E-8)
pH = 7.76
So, the correct answer is pH = 7.76, not 7.46. It seems there was a slight calculation error in your calculations.
To determine what went wrong, let's go step by step through the calculations you made.
First, you correctly identified the balanced chemical equation for the reaction:
HOCl + H2O → H3O+ + OCl-
You then determined the initial concentrations of HOCl and NaOCl:
HOCl: 0.15 M
NaOCl: 0.25 M
Next, you determined the changes in concentrations caused by bubbling HCl into the solution. You correctly subtracted 0.05 mol HCl/L from the concentration of HOCl and added 0.05 mol HCl/L to the concentration of NaOCl:
HOCl: 0.15 M - 0.05 M = 0.10 M
NaOCl: 0.25 M + 0.05 M = 0.30 M
So far, everything looks correct.
Now, let's calculate the pH using the given equation:
[H3O+] = (0.30 M)([H3O+]) / (0.10 M) = 3[H3O+]
At this point, it seems that you calculated [H3O+] to be 1.167E-8, but I believe there might have been a typo in your calculation.
Let's redo the calculation:
[H3O+] = (0.30 M)(3[H3O+]) / (0.10 M) = 9[H3O+]
Now, we need to solve for [H3O+]:
9[H3O+] = 3.5E-8
[H3O+] = (3.5E-8) / 9 ≈ 3.89E-9
To find the pH, we take the negative logarithm of [H3O+]:
pH = -log(3.89E-9) ≈ 8.41
Therefore, the correct pH of the solution should be around 8.41, not 7.78.
It seems like there was an error in calculating the concentration of H+ ions. Double-check your calculations to see if you made a mistake during the calculation process.