Buffer Solutions adding acid or base

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    Acid Base Buffer
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SUMMARY

The discussion focuses on calculating the pH and hydrogen ion concentration ([H+]) of a buffer solution created by dissolving 1.00 mole of lactic acid (HLac) and 1.00 mole of sodium lactate (NaLac) in one liter of water. The initial pH of the buffer is determined to be 2.1 using the equation pH = pKa - log[HA]/[A-]. Additionally, the discussion explores the effects of adding 1L of 0.15 M HCl and 1L of 0.15 M NaOH to the buffer, requiring further calculations for the new concentrations of [HLac], [Lac], [H+], and pH.

PREREQUISITES
  • Understanding of buffer solutions and their components
  • Familiarity with the Henderson-Hasselbalch equation
  • Knowledge of acid-base equilibria
  • Basic skills in stoichiometry and concentration calculations
NEXT STEPS
  • Calculate the new pH after adding strong acids and bases to buffer solutions
  • Study the effects of dilution on buffer capacity
  • Explore the concept of buffer range and its practical applications
  • Learn about different types of buffer solutions and their uses in biochemical contexts
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in biochemical research or applications requiring buffer solutions.

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Homework Statement


A buffer is prepared by dissolving 1.00 mole HLac (with a ka=8.0x10-3) and 1.00 mole of sodium lactate (NaLac) in enough water to form one (1) liter of solution. Calculate the [H+] and the pH of the buffer.
36 minutes ago - 3 days left to answer.
Additional Details
12 minutes ago

There is another part to this and it says
Using the information from that problem: supposed we add a strong acid or base to the buffer; calculate the following:
a. [HLac], [Lac], [H+] and the pH after adding 1L of 0.15 M-HCl
b. [HLac], [Lac], [H+], and the pH after adding 1L of 0.15M-NaOH



Homework Equations



pH=pka-log[HA]/[A-]

The Attempt at a Solution



I know the answer to the first part is found to be pH=2.1 I just have no idea how to find the part a and b for the second half.
 
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you can always try writing the equilibrium equation then solve for them using the first principles:

A + B <-> C
1 1 0
1-x 1-x x
K = x/(1-x)^2 etc
 

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