Determine volume of HCl added to propionic acid to get pH=1

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SUMMARY

The discussion focuses on calculating the volume of 1.0M HCl required to lower the pH of a 250ml solution of 0.1M propionic acid (HC3H5O2) to 1.00. The initial concentration of H+ ions was determined to be 1.16E-3, resulting in a pH of 2.93. Using the dissociation constant (Ka = 1.35E-5) and the Henderson-Hasselbalch equation, it was established that at pH 1.0, the dissociation of propionic acid is minimal, allowing for the assumption that the pH is primarily influenced by the HCl dilution. The second part of the discussion involves raising the pH to 4.00 using 1.00M NaC3H5O2 as a buffer solution.

PREREQUISITES
  • Understanding of acid-base chemistry and pH calculations
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of ICE (Initial, Change, Equilibrium) charts for dissociation
  • Basic grasp of dissociation constants (Ka) and their implications
NEXT STEPS
  • Calculate the exact volume of 1.0M HCl needed to achieve pH 1.00 using the derived equations
  • Explore the use of the Henderson-Hasselbalch equation in buffer solutions
  • Investigate the impact of different concentrations of NaC3H5O2 on pH adjustment
  • Review the principles of buffer capacity and its relevance in acid-base titrations
USEFUL FOR

Chemistry students, laboratory technicians, and educators involved in acid-base equilibrium and buffer solution studies will benefit from this discussion.

BThomas1219
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Homework Statement


Given 250ml of 0.1M HC3H5O2 (Ka=1.35E-5) What volume of
a)1.0M HCl is needed to lower the pH to 1.00.
b) 1.00M NaC3H5O2 raise pH to 4.00


Homework Equations


pH=pKa-log (HA/A)
ICE chart

The Attempt at a Solution


I used the ice chart for the dissociation of propionic acid. I determined the concentration of H+ and A- to be 1.16E-3. the pH was 2.93. I do not know how to factor in the HCl to determine the volume needed to lower the pH.
 
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K_a = \frac {[A^-][H^+]}{[HA]}

rearranged:

\frac {[A^-]}{[HA]} = 10^{pH - pK_a}

At pH 1.0 only about 101-4.75=10-3.75=2x10-4 of the acetic acid is dissociated. That means in the first case you can - more or less safely - assume presence of acetic acid doesn't matter much, as the first approximation pH will depend mostly on the HCl dilution.

Second case is just a buffer solution, use Henderson-Hasselbalch equation.
 

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