A buffer is made up of 239 mL of 0.187 M potassium hydrogen tartrate (KHC4 H4 O6) and 137 mL o

step 1 In this problem we're told that we have a buffer made up of 239 milliliters of .187 molar potass

A buffer is made up of 239 mL of 0.187 M potassium hydrogen...

A buffer is made up of $239 \mathrm{~mL}$ of $0.187 \mathrm{M}$ potassium hydrogen tartrate $\left(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right)$ and $137 \mathrm{~mL}$ of $0.288 \mathrm{M}$ potassium tartrate $\left(\mathrm{K}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right) . K_{\mathrm{a}}$ for $\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right)$ is $4.55 \times 10^{-5} .$ Assuming volumes are additive, calculate (a) the $\mathrm{pH}$ of the buffer. (b) the $\mathrm{pH}$ of the buffer after adding $0.0250 \mathrm{~mol}$ of $\mathrm{HCl}$ to $0.376 \mathrm{~L}$ of the buffer. (c) the $\mathrm{pH}$ of the buffer after adding $0.0250 \mathrm{~mol}$ of $\mathrm{KOH}$ to $0.376 \mathrm{~L}$ of the buffer.

A buffer is made up of $239 \mathrm{~mL}$ of $0.187 \mathrm{M}$ potassium hydrogen tartrate $\left(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right)$ and $137 \mathrm{~mL}$ of $0.288 \mathrm{M}$ potassium tartrate $\left(\mathrm{K}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right) . K_{\mathrm{a}}$ for $\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right)$ is $4.55 \times 10^{-5} .$
Assuming volumes are additive, calculate
(a) the $\mathrm{pH}$ of the buffer.
(b) the $\mathrm{pH}$ of the buffer after adding $0.0250 \mathrm{~mol}$ of $\mathrm{HCl}$ to $0.376 \mathrm{~L}$ of the buffer.
(c) the $\mathrm{pH}$ of the buffer after adding $0.0250 \mathrm{~mol}$ of $\mathrm{KOH}$ to $0.376 \mathrm{~L}$ of the buffer.

Key Concepts

Reaction Stoichiometry in Buffer Systems

In buffer systems, the stoichiometry of the reactions between the buffer components and any added strong acid or base must be taken into account. When a strong acid or base is added, it reacts with either the weak base or weak acid present in the buffer, altering the ratio of the conjugate pair. Proper accounting of these stoichiometric changes is essential for predicting the new pH after the addition of acid or base.

Acid-Base Equilibrium and Ka

Acid-base equilibrium involves the reversible dissociation of a weak acid into its conjugate base and hydrogen ions. The acid dissociation constant (Ka) quantifies the extent of this dissociation. Understanding Ka and its logarithmic form, pKa, is important in buffer chemistry because it helps determine the effective pH range of the buffer system and guides the calculation of pH using the Henderson-Hasselbalch equation.

Buffer Solutions

A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) and resists changes in pH when small amounts of a strong acid or base are added. This property is crucial in many chemical and biological systems where pH stability is necessary, as the components of the buffer react with any added acid or base to minimize changes in hydrogen ion concentration.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simplified formula derived from the acid dissociation equilibrium expression and is commonly used to estimate the pH of a buffer solution. It relates the pH to the pKa of the weak acid and the ratio of the concentrations of the conjugate base and the weak acid, thus providing a straightforward way to understand how changes in the buffer components affect pH.

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