Describe how you would prepare 200.0 mL of a pH=4.00 buffer solution from a solution of 0.150-M

step 1 In answering this question, let's look at the expression pH is approximately equal to PGA plus l

Describe how you would prepare 200.0 mL of a pH=4.00 buffer...

Describe how you would prepare $200.0 \mathrm{~mL}$ of a $\mathrm{pH}=4.00$ buffer solution from a solution of $0.150-\mathrm{M}$ lactic acid, $\mathrm{CH}_{3} \mathrm{CHOHCOOH}(a q)$, and a solution of $0.100-\mathrm{M}$ barium hydroxide, $\mathrm{Ba}(\mathrm{OH})_{2}(a q) .$ Take the $\mathrm{p} K_{\mathrm{a}}$ of lactic acid to be $3.86 .$

Key Concepts

Molarity and Volume Calculations

Molarity and volume calculations are fundamental when preparing solutions with a specific concentration. These calculations involve determining the number of moles of reactants required and the volumes of stock solutions necessary to achieve these moles in a final buffer solution of a given total volume. Accurate measurement ensures that the buffer has the correct ratio of acid to conjugate base as dictated by the Henderson-Hasselbalch equation.

Buffer Solution

A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in pH upon the addition of small amounts of strong acid or base. It works by neutralizing added H+ or OH– through its components, thereby maintaining the solution's pH within a narrow range. In this case, lactic acid and its conjugate base (lactate) form such a buffer system.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential tool for calculating the pH of a buffer solution. It relates the pH of the solution to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the acid. The equation, pH = pKa + log([A–]/[HA]), is used to determine the necessary proportions of acid and base to achieve a target pH value.

Acid-Base Neutralization Stoichiometry

Acid-base neutralization stoichiometry involves understanding how the moles of a weak acid and a strong base react to form the conjugate base of the acid and water. In preparing a buffer, a careful amount of strong base is added to convert a fraction of the weak acid into its conjugate base. It is crucial to account for the stoichiometric ratios in the reaction, especially when the strong base provides more than one hydroxide ion per formula unit.

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