REPOSTING: observing pH changes in Water and Buffer solutions.
the amount added of HCl and NaOH are both 25mL.
the buffer solution is made of 2grams of NaC2H3O2 and 4 mL of 6M HC2H3O2 in a 50 mL solution (46mL of water). the solution will contain 2.4x10^-2 mol each of NaC2H3O2 a d HC2H3O2.
question:
how to answer it (I'm not sure I'm doing it right and
want an expert to double check):
I need help calculating the theoretical and H3O+ concentration and the pH based off the theoretical.
1. For the distilled water, pH = -log[H+] = - log(1* 10-7) = 7
2. For your original buffer solution:
The pH of a buffer solution is calculated from the Henderson-Hasselbach equation which is given below.
where, pH = -log[H+]
pKa = -log(Ka) = -log(1.78* 10-5) =4.75 ; Ka = acid dissociation constant of the weak acid (here CH3COOH)
[salt] = concentration of the salt (here CH3COONa)
[acid] = concentration of the weak acid (here CH3COOH)
Now,
Molecular mass of CH3COONa = 82 g/mol
Therefore, 2g CH3COONa = (2/ 82) mol = 2.4* 10-2 mol CH3COONa.
[CH3COOH] is calculated from the dilution principle, which is
where,
Therefore,
3. Distilled water with added 0.5 mL 6M HCl solution:
Since HCl is a strong acid, it will be completely dissociated and produce a solution of 6M H+ solution. Now, the contribution of water to the concentration of H+ is negligible compared to that produced by HCl.
We will again use the formula to calculate the final concentration of H+ in the solution.
Therefore,
4. For bufffer solution+0.5 mL 6M HCl solution:
0.5 mL 6M HCl contains of HCl
This will convert the 3* 10-3 moles of CH3COONa to form additional 3* 10-3 moles of CH3COOH.
Therefore, the concentrations after adding HCl are
50 mL of 0.48M CH3COOH contains 2.4* 10-2 CH3COOH.
Therefore,
5.Distilled water with added 0.5 mL 6M NaOH solution:
Since NaOH is a strong base, it will be completely dissociated and produce a solution of 6M OH- .
We will again use the formula to calculate the final concentration of OH- in the solution.
Therefore,
6.For bufffer solution+0.5 mL 6M NaOH solution:
0.5 mL 6M NaOH contains of NaOH.
This will convert the 3* 10-3 moles of CH3COOH to form additional 3* 10-3 moles of CH3COONa.
Therefore, the concentrations after adding NaOH are
50 mL of 0.48M CH3COOH contains 2.4* 10-2 CH3COOH.
Therefore,
REPOSTING: observing pH changes in Water and Buffer solutions. the amount added of HCl and NaOH...
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