/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 Calculate the pH of a buffer sol... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH \(_{2} \mathrm{PO}_{4}\), and enough water to make 0.500 L of solution.

Short Answer

Expert verified
The pH of the buffer solution is 2.15 + log(0.500/0.310), which needs to be calculated for the exact pH value.

Step by step solution

01

Identify the Acid and Conjugate Base

First, identify the components of the buffer solution. Here, phosphoric acid (H_3PO_4) acts as the weak acid and potassium dihydrogen phosphate (KH_2PO_4) supplies the dihydrogen phosphate ions (H_2PO_4^-), which act as the conjugate base.
02

Write the Equilibrium Expression

Write the acid dissociation reaction for phosphoric acid and the expression for its equilibrium constant (K_a): H_3PO_4 ↔ H^+ + H_2PO_4^-. The equilibrium expression is K_a = [H^+][H_2PO_4^-]/[H_3PO_4].
03

Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for a buffer is pH = pK_a + log([A^-]/[HA]), where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the acid. To compute the pH, we need the concentration of H_2PO_4^- and H_3PO_4 and the pK_a value for the dissociation of H_3PO_4 into H_2PO_4^-.
04

Calculate the Concentrations

Find the molar concentrations by dividing the number of moles by the total volume of the solution in liters. For H_2PO_4^-, the concentration is 0.250mol / 0.500L = 0.500M. For H_3PO_4, it's 0.155mol / 0.500L = 0.310M.
05

Find the pKa Value

Look up the pKa value for phosphoric acid. For this exercise, we use the appropriate pKa value for the first dissociation, which is typically around 2.15.
06

Apply the Henderson-Hasselbalch Equation

Plug in the values into the Henderson-Hasselbalch equation: pH = 2.15 + log(0.500/0.310). Calculate the log value and add to the pKa to find the pH.
07

Calculate the pH

Finish the calculation to find the final pH value of the buffer solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch Equation
Understanding the Henderson-Hasselbalch equation is crucial when calculating the pH of buffer solutions. A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

The Henderson-Hasselbalch equation offers a straightforward way to estimate the pH of a buffer based on the concentrations of these components. It is expressed as: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] where \( \text{p}K_a \) is the negative logarithm of the acid dissociation constant (Ka), \( [\text{A}^-] \) is the concentration of the conjugate base, and \( [\text{HA}] \) is the concentration of the acid. In the given exercise, by applying the Henderson-Hasselbalch equation, students can easily calculate the pH of the buffer solution using the known values of phosphoric acid and its conjugate base concentration.
Buffer Solution Chemistry
Buffer Solution Chemistry involves solutions that resist changes in pH upon the addition of small amounts of strong acid or base. These solutions are composed of a weak acid or base and its conjugate counterpart.

In the context of the exercise, phosphoric acid and potassium dihydrogen phosphate form a buffer system. The phosphoric acid, \(H_3PO_4\), is the weak acid, and the potassium dihydrogen phosphate, \(KH_2PO_4\), provides the conjugate base, \(H_2PO_4^-\). These components work together to maintain pH stability. The effectiveness of a buffer depends on the concentration ratio between the acid and conjugate base, and the absolute concentrations of buffer components.
Acid-Base Equilibrium
Acid-Base Equilibrium describes the state of balance between dissociated (ions) and undissociated forms of acids and bases in a solution. For weak acids, this is represented by the equation:\[ \text{HA} \leftrightarrow \text{H}^+ + \text{A}^- \]

The position of equilibrium is determined by the acid dissociation constant (Ka), which reflects the strength of the acid. A smaller Ka indicates a weaker acid. Coming back to our exercise, we focus on the first dissociation of phosphoric acid, which establishes an equilibrium between undissociated phosphoric acid, hydrogen ions, and dihydrogen phosphate ions. By understanding this equilibrium, students can appreciate why the ratio of conjugate base to acid (\(\frac{[\text{A}^-]}{[\text{HA}]}\)) is so important in the Henderson-Hasselbalch equation.
Acid Dissociation Constant (Ka)
The Acid Dissociation Constant (Ka) quantifies the strength of an acid in solution. It is defined by the equilibrium constant for the reaction of the acid (HA) dissociating into hydrogen ions (H+) and its conjugate base (A-):\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]

For weak acids, the lower the Ka value, the less the acid dissociates, which means a weaker acid. The pKa value is simply the negative logarithm of Ka: \( \text{p}K_a = -\log(K_a) \).In practical terms, as in the given problem, the pKa value is used directly in the Henderson-Hasselbalch equation to calculate pH levels of buffer solutions. The pKa for the first dissociation of phosphoric acid is around 2.15, which helps determine the acid's relative strength and, consequently, the buffer's pH.

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Most popular questions from this chapter

Predict which acid in each of the following pairs is the stronger and explain your reasoning for each. (a) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{HF}\) (b) \(\mathrm{B}(\mathrm{OH})_{3}\) or \(\mathrm{Al}(\mathrm{OH})_{3}\) (c) \(\mathrm{HSO}_{3}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) (d) \(\mathrm{NH}_{3}\) or \(\mathrm{H}_{2} \mathrm{S}\) (e) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{Te}\)

State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species: (a) \(\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (c) \(\mathrm{S}^{2-}\) (d) \(\mathrm{CO}_{3}^{2-}\) (e) \(\mathrm{HSO}_{4}^{-}\)

Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid ( \(K_{\mathrm{a}}=9.8\) \(\left.\times 10^{-5}\right)\) with \(0.100 \mathrm{M} \mathrm{KOH}\). (a) no KOH added (b) \(20 \mathrm{mL}\) of KOH solution added (c) \(39 \mathrm{mL}\) of \(\mathrm{KOH}\) solution added (d) \(40 \mathrm{mL}\) of \(\mathrm{KOH}\) solution added (e) \(41 \mathrm{mL}\) of KOH solution added

Calculate the concentration of each species present in a \(0.010-M\) solution of phthalic acid, \(\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}\). $$\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)\left(\mathrm{CO}_{2}\right)^{-}(a q) \quad K_{\mathrm{a}}=1.1 \times 10^{-3}$$, $$\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2} \mathrm{H}\right)\left(\mathrm{CO}_{2}\right)(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CO}_{2}\right)_{2}^{2-}(a q) \quad K_{\mathrm{a}}=3.9 \times 10^{-6}$$.

Salicylic acid, HOC \(_{6} \mathrm{H}_{4} \mathrm{CO}_{2} \mathrm{H}\), and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838. (a) Both functional groups of salicylic acid ionize in water, with \(K_{\mathrm{a}}=1.0 \times 10^{-3}\) for the \(-\mathrm{CO}_{2} \mathrm{H}\) group and 4.2 \(\times 10^{-13}\) for the \(-\mathrm{OH}\) group. What is the \(\mathrm{pH}\) of a saturated solution of the acid (solubility \(=1.8 \mathrm{g} / \mathrm{L}\) ). (b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{CO}_{2} \mathrm{H} .\) The \(-\mathrm{CO}_{2} \mathrm{H}\) functional group is still present, but its acidity is reduced, \(K_{\mathrm{a}}=3.0 \times 10^{-4} .\) What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).

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