Question

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 pH? (Assume that volumes are additive.)

Short answer

Question: Calculate the ratio of [HCO3-]/[CO32-] and the amounts of sodium hydrogen carbonate and sodium carbonate required to prepare a buffer with a pH of 9.40. Additionally, determine the volume of sodium hydrogen carbonate solution needed. Answer: The desired ratio of [HCO3-]/[CO32-] is approximately 0.1166. To prepare the buffer, we need 0.02624 moles of sodium hydrogen carbonate and 145.0 grams of sodium carbonate. The volume of 0.200 M NaHCO3 solution needed to prepare the buffer is 59.5 mL.

Step-by-step solution

To calculate the desired ratio of [HCO3-]/[CO32-], we will use the Henderson-Hasselbalch equation: \( \mathrm{pH} = \mathrm{pK_a} + \log \frac{\left[\mathrm{A}^-\right]}{\left[\mathrm{HA}\right]}\) where \(\mathrm{pH}\) is the desired acidity of the buffer, \(\mathrm{pK_a}\) is the negative logarithm of the acid dissociation constant, \(\mathrm{K_a}\), for the conjugate acid (in this case, it is \(\mathrm{HCO}_3^-\)), [A-] represents the concentration of the conjugate base (in this case, \(\mathrm{CO}_3^{2-}\)), and [HA] represents the concentration of the conjugate acid (in this case, \(\mathrm{HCO}_3^-\)). First, we need to determine the \(\mathrm{pK_a}\) for the reaction. The \(\mathrm{K_a}\) for the reaction is \(4.67 \times 10^{-11}\). Thus, the \(\mathrm{pK_a}\) is \(\mathrm{pK_a} = -\log \left(4.67 \times 10^{-11}\right) = 10.33\) Now we plug in the given pH and the calculated pKa into the Henderson-Hasselbalch equation: \(9.40 = 10.33 + \log \frac{\left[\mathrm{CO}_3^{2-}\right]}{\left[\mathrm{HCO}_3^-\right]} \) Solve for the ratio: \(\log \frac{\left[\mathrm{CO}_3^{2-}\right]}{\left[\mathrm{HCO}_3^-\right]} = -0.93 \) \(\frac{\left[\mathrm{CO}_3^{2-}\right]}{\left[\mathrm{HCO}_3^-\right]} = 10^{-0.93} = 0.1166\) The desired ratio of [HCO3-]/[CO32-] is approximately 0.1166. #b)# Amount of sodium hydrogen carbonate required

To calculate the moles of sodium hydrogen carbonate needed, we first need to determine the concentration of CO32- from the Na2CO3 solution: \([\mathrm{CO}_3^{2-}] = 0.225\,\mathrm{M}\) Now, we can use the determined ratio 0.1166 to find the concentration of HCO3-: \([\mathrm{HCO}_3^{-}] = [\mathrm{CO}_3^{2-}] \times 0.1166\) \([\mathrm{HCO}_3^{-}] = 0.225 \, \mathrm{M} \times 0.1166 = 0.02624 \,\mathrm{M}\) Finally, the moles of sodium hydrogen carbonate required in 1 L of solution can be determined by multiplying the concentration by the volume: \(\mathrm{moles\:of\:NaHCO_3} = 0.02624 \,\mathrm{M} \times 1\,\mathrm{L} = 0.02624\,\mathrm{mol}\) We need 0.02624 moles of sodium hydrogen carbonate to prepare the buffer. #c)# Amount of sodium carbonate required

First, determine the concentration of HCO3- from the NaHCO3 solution: \([\mathrm{HCO}_3^{-}] = 0.336\,\mathrm{M}\) Next, use the desired ratio to find the concentration of CO32-: \([\mathrm{CO}_3^{2-}] = \frac{[\mathrm{HCO}_3^{-}]}{0.1166}\) \([\mathrm{CO}_3^{2-}] = \frac{0.336\,M}{0.1166} = 2.882\,\mathrm{M}\) Now, we will calculate the moles of CO32- required in 475 mL: \(\mathrm{moles\:of\:CO}_3^{2-} = 2.882\,\mathrm{M} \times 0.475\,\mathrm{L} = 1.369\,\mathrm{mol}\) Lastly, we need to convert moles to grams, using the molar mass of sodium carbonate (Na2CO3): \(\mathrm{grams\:of\:Na}_2\mathrm{CO}_3 = 1.369\, \mathrm{mol} \times 105.99\,\frac{\mathrm{g}}{\mathrm{mol}} = 145.0\, \mathrm{g}\) We need 145.0 grams of sodium carbonate to prepare the buffer. #d)# Volume of sodium hydrogen carbonate solution required

First, determine the concentration of CO32- from the Na2CO3 solution: \([\mathrm{CO}_3^{2-}] = 0.139\,\mathrm{M}\) Next, use the desired ratio to find the concentration of HCO3-: \([\mathrm{HCO}_3^{-}] = [\mathrm{CO}_3^{2-}] \times 0.1166\) \([\mathrm{HCO}_3^{-}] = 0.139 \, \mathrm{M} \times 0.1166 = 0.01621 \,\mathrm{M}\) Now we can calculate the moles of HCO3- required using the initial volume of Na2CO3: \(\mathrm{moles\:of\:HCO}_3^{-} = 0.01621\,\mathrm{M} \times 0.735\,\mathrm{L} = 0.0119\,\mathrm{mol}\) To calculate the volume of 0.200 M NaHCO3 needed to provide this amount, we will use the moles of HCO3- and the given concentration of the NaHCO3 solution: \(\mathrm{volume\:of\:NaHCO_3\:solution} = \frac{0.0119\,\mathrm{mol}}{0.200\,\mathrm{M}} = 0.0595\,\mathrm{L}\) We need 59.5 mL of 0.200 M NaHCO3 solution to prepare the buffer.

Key concepts

Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a key element in buffer solution chemistry. It allows us to calculate the pH of a buffer solution based on the concentrations of an acid and its conjugate base.

In a buffer comprising a weak acid or base and its salt, the equation is given by:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{\left[ \text{A}^- \right]}{\left[ \text{HA} \right]} \right) \]
Here, \( \text{pH} \) is the measure of acidity, \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant (\( \text{K}_a \)) for the conjugate acid, \( \left[ \text{A}^- \right] \) is the concentration of the conjugate base, and \( \left[ \text{HA} \right] \) is the concentration of the acid. This equation is pivotal because it connects the measurable pH to the ratio of concentrations in the buffer, illuminating how changes in this ratio affect the pH.

For instance, if we're given a pH and we need to find the optimal concentration ratio of components in a buffer solution, like the sodium hydrogen carbonate-sodium carbonate system, we can rearrange the equation solving for the ratio. This brings a clearer insight into how buffers resist changes in pH when acids or bases are added to the solution.

Acid-Base Equilibrium in Buffer Solutions

The acid-base equilibrium is crucial to maintaining the proper functioning of buffer solutions. These solutions are mixtures that consist of a weak acid and its conjugate base (or a weak base and its conjugate acid), and their primary role is to resist changes in pH when small amounts of additional acids or bases are introduced.

The equilibrium in a buffer system occurs because the weak acid (HA) can donate protons (H+) to neutralize added bases, while the conjugate base (A-) can absorb protons to neutralize added acids.
\[\text{HA} \rightleftharpoons \text{A}^- + \text{H}^+\]
The principle of Le Châtelier's predicts that adding more of a product or reactant to the system will shift the equilibrium to counteract the change, enabling the buffer to 'absorb' the extra ions without a significant change in pH.

Understanding this equilibrium is fundamental when solving problems like determining the necessary components and quantities for a buffer with a specific pH. Also, acknowledging that the buffer capacity has limits and that the buffering action ceases once the weak acid or base is fully consumed helps avoid mistakes when designing or using buffer solutions.

Molarity Calculations in Buffer Preparation

Molarity (M) is a term that represents the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of solution in liters. When preparing buffer solutions, accurate molarity calculations are essential for achieving the desired pH.

The process of molarity calculation usually involves determining the number of moles of each buffer component (the weak acid and its conjugate base) and then converting these to the masses or volumes required.
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Moreover, when combining solutions that already contain the buffer components, as in the sodium hydrogen carbonate and sodium carbonate buffer system, it's important to consider the existing molarities to calculate the correct volumes or masses needed to reach the target pH.

Utilizing molarity calculations enables the precise formulation of buffer solutions, which is critical in various fields, further underlying their importance. Whether you're a student in a chemistry lab or a professional in a research setting, mastery of molarity calculations is an indispensable tool for success.