Solving a Buffer Problem in Chemistry: pH Calculation

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SUMMARY

The discussion focuses on calculating the pH of a buffer solution consisting of 0.500 L of 0.300 M KH2PO4 and 0.400 M K2HPO4 after the addition of 0.0500 moles of HCl and NaOH. The relevant formula used is pH = pKa + log([B-]/[HB]), where the weak base is K2HPO4 and the conjugate acid is KH2PO4. The common ion effect is acknowledged, indicating that the concentrations of the buffer components remain the same at equilibrium. The calculations require determining the new ratios of the buffer components post-reaction with the strong acids and bases.

PREREQUISITES
  • Understanding of buffer solutions and their components
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of strong acid and strong base reactions with weak acids and bases
  • Basic skills in stoichiometry and equilibrium concepts
NEXT STEPS
  • Study the Henderson-Hasselbalch equation in detail
  • Learn about the common ion effect in buffer solutions
  • Explore the impact of strong acids and bases on buffer capacity
  • Practice buffer pH calculations with different concentrations and volumes
USEFUL FOR

Chemistry students, educators, and professionals involved in analytical chemistry or laboratory work who need to understand buffer systems and pH calculations.

gigi9
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here is a buffer problem in my HW that I'm not sure how to do.
Plz help me. Plz show me how. Thanks a lot.
Formula: pH = pKa + log([B-]/[HB])
***A buffer is made up of 0.500 L each of .300 M KH2PO4 and .400 M K2HPO4. Calculate
a) The pH of the buffer after the addition of 0.0500 mole of HCl
b) The pH of the buffer after the addition of 0.0500 mole of NaOH.
 
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Due to the common ion effect the concentrations of the buffer should be the same at equilibrium. Use the equation to calculate the pH before the addition of any strong acid or strong base.

The acid will react with the weak base. Figure out which component of the buffer is the base. After the mole to mole reaction, less base buffer component will be left and more of the conjugate acid will form. Write out an equation to see this for yourself. Use this new ratio and substitute into the equation.

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http://groups.msn.com/GeneralChemistryHomework
 
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