Chemistry) Buffers and Ph differences question

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SUMMARY

The discussion focuses on preparing a buffer solution with a pH of 4.50 using 100 mL of 0.12 mol/L acetic acid and 0.10 mol/L NaOH. To achieve this, the Henderson-Hasselbalch equation is applied to determine the correct ratio of acetic acid to sodium acetate. Additionally, to raise the pH to 5.1, the amount of 0.20 mol/L NaOH required is calculated. The impact of adding the same amount of NaOH to 250 mL of deionized water is also analyzed, demonstrating the resulting pH change.

PREREQUISITES
  • Understanding of buffer solutions and their components
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of molarity and solution preparation
  • Basic acid-base chemistry principles
NEXT STEPS
  • Study the Henderson-Hasselbalch equation in detail
  • Learn about buffer capacity and its implications in chemical solutions
  • Explore the effects of different concentrations of NaOH on pH levels
  • Investigate the preparation and properties of sodium acetate as a buffer component
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Chemistry students, laboratory technicians, and educators seeking to deepen their understanding of buffer solutions and pH adjustments in chemical experiments.

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Homework Statement



A Chemist wishes to prepare 250mL of buffer that has a Ph of 4.50. Starting with 100mL of 0.12 mol L acetic acid and a supply of 0.10 mol L NaOh.
1) explain how this can be done
3) how much 0.20 molL NaOH must e added to this buffer to raise the pH to 5.1?
3) if the same amount of 0.20 mol L NaOH were added to 250mL of Dionized water, what would the new pH be?

Homework Equations



I know the henderson equation has to be used somewhere but I'm honestly stumped at where to even start :blushing: any general push in the right direction would be seriously appreciated, It's a bonus Lab question, but I'd still like to get any extra marks I could :)
 
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When you add NaOH to the solution of acetic acid, you neutralize it, producing sodium acetate. Just add enough to get a correct ratio given by the HH equation.
 

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