A solution of volume 80.0 mL contains 16.0 mmol HCHO2 and 9.00 mmol NaCHO2.
If 1.00 ml of 12 M HCl is added to this, what will be the resulting pH?
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
First, however, you need to make some adjustments to the acid/base because you have added some HCl to it.
HCOOH = 16.0 mmole initially.
HCOONa = 9.00 mmol initially. Adding 1.00 mL of 12 M HCl adds 12 x 1 = 12 mmole (is that 12 or 12.0 or 12.00 mmole?) to this base. Therefore, 9.00 + 12.0 = 21.0 mmole base.
pH = pKa + log[(21.0)/(16.0)] = ??
If your prof is picky, s/he will not like this BECAUSE (base) and (acid) are concentrations and not mmols. Technically, then, the concn is mmols/mL = 21.0/81.0 for (base) and 16.00/81.0 for (acid) so the equation is
pH = pKa + log[(21.0/81.0)/(16.0/81.0)] = ??. You will note that the 81.0 mL volume cancels and some profs just don't put it there. I ALWAYS counted off for not including the volume; however, since it ALWAYS cancels, I would allow the student to use V as in
pH = pKa + log [(21.0/v)/(16.0/v)] = ??. That way the student let me know that a volume went there but it would cancel and never entered into the calculation.
Sorry, there's one thing I don't understand. If an acid is being added to a buffer solution, shouldn't the pH decrease? By following your method, the answer I'm getting is greater than the original pH.
Buffers increase ph of acids, they buffer the degree at which acids affect solutions
To determine the resulting pH of the solution, we need to follow these steps:
Step 1: Calculate the total molar concentration of HCHO2 and NaCHO2.
To do this, we convert the volumes (in mL) to liters:
- HCHO2 volume = 16.0 mmol / 1000 mL/L = 0.016 mol/L
- NaCHO2 volume = 9.00 mmol / 1000 mL/L = 0.009 mol/L
Step 2: Calculate the total moles of HCHO2 and NaCHO2.
- Total moles = moles of HCHO2 + moles of NaCHO2
- Total moles = 0.016 mol/L + 0.009 mol/L = 0.025 mol/L
Step 3: Calculate the total volume of the solution after adding 1.00 mL of 12 M HCl.
- Total volume = initial volume + volume of HCl added
- Total volume = 80.0 mL + 1.00 mL = 81.0 mL
Step 4: Calculate the final molar concentration of the solutions after dilution.
- Final concentration = total moles / total volume
- Final concentration = 0.025 mol / 0.081 L = 0.308 M
Step 5: Determine the initial concentration of HCl after dilution.
- Initial concentration of HCl = (volume of HCl added * concentration of HCl added) / total volume
- Initial concentration of HCl = (1.00 mL * 12 M) / 0.081 L = 148.148 M
Step 6: Calculate the moles of HCl present in the solution.
- Moles of HCl = initial concentration of HCl * total volume
- Moles of HCl = 148.148 M * 0.081 L = 11.999 moles
Step 7: Calculate the final concentration of HCHO2 and NaCHO2.
- Final concentration of HCHO2 = initial concentration of HCHO2 - moles of HCl
- Final concentration of HCHO2 = 0.308 M - 11.999 moles = -11.691 M (concentration would become negative, which is not physically possible)
- Final concentration of NaCHO2 = initial concentration of NaCHO2 - moles of HCl
- Final concentration of NaCHO2 = 0.308 M - 11.999 moles = -11.691 M (concentration would become negative, which is not physically possible)
Step 8: Calculate the pH of the resulting solution using the concentrations of H+ ions.
Since the concentrations of HCHO2 and NaCHO2 would become negative after adding HCl, we cannot directly calculate the pH using these concentrations. Further information or clarification of the problem is needed to proceed with the calculation.
Please double-check the information provided and ensure all the values are accurate.