The real question I have is that will there be an effect on pH by the water in the dilute solutions?
There is very little effect on the pH of a buffer when you dilute it with water (as long as the concentration is still high compared to that of hydronium and hydroxide concentrations).
I need some direction in being able to figure out the pH of the diluted solution and how to calculate the effect of .1 mL 1.25 M HCl addition to each of the dilute solutions.
The buffer capacity is lowered by diluting the buffer. If you add the same amount of strong acid to a buffer or a diluted buffer, the diluted buffer will show a larger change in pH.
If I understand the calculations presented in the question correctly, the concentrations are off by a factor 10. Comparing the 150 mM with the 30 mM buffer, the relative change in concentrations upon adding the same amount of HCl is larger for the more diluted buffer, thus the pH change is predicted to be more pronounced. The experimental data supports that prediction.
Here are the updated calculations for row 2, 150 mM:
For the addition of .1 mL 1.25 M HCl, we can try the same approach:
pH $\displaystyle = 8.3 + \log \frac{\pu{0.0075 M} - \pu{0.0125 M}}{\pu{0.0225 M} + \pu{0.0125 M}} = $ oops, we ran out of buffer (HCl is in excess of the deprotonated form of the buffer substance).
The real question I have is that will there be an effect on pH by the water in the dilute solutions?
There is very little effect on the pH of a buffer when you dilute it with water (as long as the concentration is still high compared to that of hydronium and hydroxide concentrations).
I need some direction in being able to figure out the pH of the diluted solution and how to calculate the effect of .1 mL 1.25 M HCl addition to each of the dilute solutions.
The buffer capacity is lowered by diluting the buffer. If you add the same amount of strong acid to a buffer or a diluted buffer, the diluted buffer will show a larger change in pH.
If I understand the calculations presented in the question correctly, the concentrations are off by a factor 10. Comparing the 150 mM with the 30 mM buffer, the relative change in concentrations upon adding the same amount of HCl is larger for the more diluted buffer, thus the pH change is predicted to be more pronounced. The experimental data supports that prediction.
Here are the updated calculations for row 2, 150 mM:
For the addition of .1 mL 1.25 M HCl, we can try the same approach:
pH $\displaystyle = 8.3 + \log \frac{\pu{0.0075 M} - \pu{0.0125 M}}{\pu{0.0225 M} + \pu{0.0125 M}} = $ oops, we ran out of buffer (HCl is in excess of the deprotonated form of the buffer substance).
Hmm, with different charges of respective conjugate acid and base, their activities would change by dilution at different rate. I would have to quantify it, but I am not sure if it is very little for different ionic strengths. OTOH, if the ionic strength is kept constant by added salts, then you are right.More
There is very little effect on the pH of a buffer when you dilute it with water (as long as the concentration is still high compared to that of hydronium and hydroxide concentrations).
The buffer capacity is lowered by diluting the buffer. If you add the same amount of strong acid to a buffer or a diluted buffer, the diluted buffer will show a larger change in pH.
If I understand the calculations presented in the question correctly, the concentrations are off by a factor 10. Comparing the 150 mM with the 30 mM buffer, the relative change in concentrations upon adding the same amount of HCl is larger for the more diluted buffer, thus the pH change is predicted to be more pronounced. The experimental data supports that prediction.
Here are the updated calculations for row 2, 150 mM:
pH $\displaystyle = 8.3 + \log\frac{\pu{0.0375 M} }{\pu{0.1125 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl:
pH $\displaystyle = 8.3 + \log \frac{\pu{0.0375 M} - \pu{0.0125 M}}{\pu{0.1125 M} + \pu{0.0125 M}} = 7.50$
For the 30 mM buffer, the buffer concentration is one fifth of the 150 mM buffer, but the amount (and concentration) of HCl added is the same:
pH $\displaystyle = 8.3 + \log\frac{\pu{0.0075 M} }{\pu{0.0225 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl, we can try the same approach:
pH $\displaystyle = 8.3 + \log \frac{\pu{0.0075 M} - \pu{0.0125 M}}{\pu{0.0225 M} + \pu{0.0125 M}} = $ oops, we ran out of buffer (HCl is in excess of the deprotonated form of the buffer substance).
pH $= \log (0.0125 - 0.0075) = \log 0.005 = 2.3$
This matches the experimental result fairly well.
There is very little effect on the pH of a buffer when you dilute it with water (as long as the concentration is still high compared to that of hydronium and hydroxide concentrations).
The buffer capacity is lowered by diluting the buffer. If you add the same amount of strong acid to a buffer or a diluted buffer, the diluted buffer will show a larger change in pH.
If I understand the calculations presented in the question correctly, the concentrations are off by a factor 10. Comparing the 150 mM with the 30 mM buffer, the relative change in concentrations upon adding the same amount of HCl is larger for the more diluted buffer, thus the pH change is predicted to be more pronounced. The experimental data supports that prediction.
Here are the updated calculations for row 2, 150 mM:
pH $\displaystyle = 8.3 + \log\frac{\pu{0.0375 M} }{\pu{0.1125 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl:
pH $\displaystyle = 8.3 + \log \frac{\pu{0.0375 M} - \pu{0.0125 M}}{\pu{0.1125 M} + \pu{0.0125 M}} = 7.50$
For the 30 mM buffer, the buffer concentration is one fifth of the 150 mM buffer, but the amount (and concentration) of HCl added is the same:
pH $\displaystyle = 8.3 + \log\frac{\pu{0.0075 M} }{\pu{0.0225 M}} = 7.822$
For the addition of .1 mL 1.25 M HCl, we can try the same approach:
pH $\displaystyle = 8.3 + \log \frac{\pu{0.0075 M} - \pu{0.0125 M}}{\pu{0.0225 M} + \pu{0.0125 M}} = $ oops, we ran out of buffer (HCl is in excess of the deprotonated form of the buffer substance).
pH $= \log (0.0125 - 0.0075) = \log 0.005 = 2.3$
This matches the experimental result fairly well.
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