A sodium hydrogen carbonate-sodium carbonate buffer is to be...

A sodium hydrogen carbonate-sodium carbonate buffer is to be prepared with a pH of 9.40 . (a) What must the $\left[\mathrm{HCO}_{3}^{-}\right] /\left[\mathrm{CO}_{3}^{2-}\right]$ ratio be? (b) How many moles of sodium hydrogen carbonate must be added to a liter of $0.225 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ to give this $\mathrm{pH}$ ? (c) How many grams of sodium carbonate must be added to $475 \mathrm{~mL}$ of $0.336 \mathrm{M} \mathrm{NaHCO}_{3}$ to give this $\mathrm{pH}$ ? (Assume no volume change.) (d) What volume of $0.200 \mathrm{M} \mathrm{NaHCO}_{3}$ must be added to $735 \mathrm{~mL}$ of a $0.139 \mathrm{M}$ solution of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ to give this $\mathrm{pH}$ ? (Assume that volumes are additive.)

A sodium hydrogen carbonate-sodium carbonate buffer is to be prepared with a pH of 9.40 .
(a) What must the $\left[\mathrm{HCO}_{3}^{-}\right] /\left[\mathrm{CO}_{3}^{2-}\right]$ ratio be?
(b) How many moles of sodium hydrogen carbonate must be added to a liter of $0.225 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}$ to give this $\mathrm{pH}$ ?
(c) How many grams of sodium carbonate must be added to $475 \mathrm{~mL}$ of $0.336 \mathrm{M} \mathrm{NaHCO}_{3}$ to give this $\mathrm{pH}$ ? (Assume no volume change.)
(d) What volume of $0.200 \mathrm{M} \mathrm{NaHCO}_{3}$ must be added to $735 \mathrm{~mL}$ of a $0.139 \mathrm{M}$ solution of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ to give this $\mathrm{pH}$ ? (Assume that volumes are additive.)

Key Concepts

Molarity and Volume Relationships

Molarity, defined as the number of moles of a solute per liter of solution, is a core concept for quantifying the amounts of chemical species in solution. Coupled with volume measurements, this concept allows for the calculation of the exact number of moles present, which is critical when preparing solutions with precise pH requirements or carrying out titrations and other stoichiometric calculations.

Acid–Base Equilibria

Acid–base equilibria involve the reversible exchange of protons (H?) between species. Understanding these equilibria is essential for predicting the behavior of conjugate acid–base pairs in solution. This concept explains how a weak acid and its conjugate base (or vice versa) establish an equilibrium that can be manipulated to maintain a specific pH in buffer systems.

Stoichiometric Calculations

Stoichiometric calculations involve using the relationships between the amounts of reactants and products in chemical reactions. In the context of preparing buffer solutions, these calculations are used to determine how many moles (or grams) of a given component must be added to achieve a specific ratio of components, thereby controlling the pH. This requires balancing the chemical equation and applying the concept of molarity and volume to achieve the desired composition.

Buffer Solutions

Buffer solutions are mixtures of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resist changes in pH upon the addition of small amounts of acid or base. In the context of acid–base chemistry, understanding how buffers work is crucial for controlling the pH of a solution in various chemical and biological systems.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simplified version of the acid dissociation expression that relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentrations of the conjugate base and the acid. It is especially useful for designing buffer solutions by allowing the calculation of the necessary proportions of acid and conjugate base to achieve a desired pH.

Transcript

00:01 Okay, so we have a buffer solution, and we're going to have some hydrogen carbonate as the weak acid and some carbonate as the weak base.

00:09 So we know the ph is given as 9 .40.

00:14 So let's solve for the h -plus concentration.

00:19 So that's 10 to the negative 9 .40 or 3 .98 times 10 to negative 10 molar.

00:32 Our buffer, our weak acid is the bicarbonate, and our weak base is the carbonate.

00:38 So we're going to need the ka for bicarbonate.

00:41 So that will be 4 .7 times 10 to negative 11.

00:49 So our equation for buffers is h plus is ka times the concentration of the acid over the concentration of the base.

01:01 So we know the h plus and the ka, so we can find that ratio of acid to base concentration.

01:10 So the h plus is 3 .98 times 10 to the negative 10.

01:17 We'll set that equal to our ka.

01:22 And again, then that's just multiplied by this ratio of the concentration of the acid over the concentration of the base.

01:32 So the concentration over acid, which is bicarbonate, divided by the base.

01:41 Concentration of carbonate.

01:45 We'll just divide that out and we'll get a ratio of 8 .47.

01:53 Okay, so we've got our ratio.

01:55 If that ratio equals the concentration, since concentration is moles per liter over moles per liter, and the volumes of course have to be the same.

02:08 So we can also say that the ratio is the moles of acid over the moles of base will equal 8 .47.

02:20 So you can do it either, in moles or in molarity...

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