Could someone explain how to do this? For the next three problems, consider 1.0 L of a solution...

Question:

Could someone explain how to do this? For the next three problems, consider 1.0 L of a solution which is{eq}0.35 M HC_2H_3O_2 and 0.2 M NaC_2H_3CO_2 (K_a for HC_2H_3O_2 = 1.8 times 10^-5) {/eq}. Assume 2 significant figures in all of the given concentrations so that you should calculate all of the following pH values to two decimal places. Calculate the pH of this solution. Calculate the pH after 0.10 mol of HCl has been added to the original solution. Assume no volume change on addition of HCl. Calculate the pH after 0.20 mol of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Henderson-Hasselbalch Equation

One can use the Henderson-Hasselbalch equation to predict the pH of a buffer solution. Similarly, the equation can be used to calculate the recipe needed to make a buffer of the desired solution. The Henderson-Hasselbalch equation is

$$pH = pK_a + log \frac {[A^-]}{[HA]} $$

Answer and Explanation:

  • a typo in the sodium salt (extra carbon) is ignored in the following answer.

The pH of the original solution is a buffer because it contains both a weak acid and its conjugate base. Its pH can be calculated from the Henderson-Hasselbalch equation. Sodium ion is ignored because it is pH neutral.

$$pH = pK_a + log \frac { [A^-] }{ [HA] }\\ pH = pK_a + log \frac { [C_2H_3O_2^-] }{ [HC_2H_3CO_2] } = -\log (1.8 \times 10^{-5} ) + log \frac {0.2 \ M}{0.35 \ M} =\boxed {4.50} $$


After addition of 0.10 mole of HCl, some of the conjugate base will react with it and form more of the weak acid.

$$C_2H_3O_2^- + H^+ \rightarrow HC_2H_3O_2 $$

The new equilibrium concentrations are

$$[C_2H_3O_2^-] = (0.2 - 0.10) \ M \\ [HC_2H_3O_2] = (0.35 + 0.10) \ M $$

The new pH is calculated

$$pH = pK_a + log \frac { [C_2H_3O_2^-] }{ [HC_2H_3CO_2] } = -\log (1.8 \times 10^{-5} ) + log \frac {0.1 \ M}{0.45 \ M} =\boxed {4.09} $$

As expected, adding HCl lowers the pH.


After addition of 0.20 mole of NaOH, some of the acid will react with it and form more of the conjugate base

$$HC_2H_3O_2 + NaOH \rightarrow C_2H_3O_2^- $$

The new equilibrium concentrations are

$$[C_2H_3O_2^-] = (0.2 + 0.20) \ M \\ [HC_2H_3O_2] = (0.35 - 0.20) \ M $$

The new pH is calculated

$$pH = pK_a + log \frac { [C_2H_3O_2^-] }{ [HC_2H_3CO_2] } = -\log (1.8 \times 10^{-5} ) + log \frac {0.40 \ M}{0.15 \ M} =\boxed {5.17} $$

As expected, adding NaOH increases the pH.


Learn more about this topic:

The Henderson Equation: Definition & Examples
The Henderson Equation: Definition & Examples

from National Eligibility Test (AIPMT): Study Guide

Chapter 24 / Lesson 4
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