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Chapter 17: Problem 12

What is the \(\mathrm{pH}\) of the buffer \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} / 0.15 \mathrm{M}\) \(\mathrm{KH}_{2} \mathrm{PO}_{4} ?\)

Short Answer

Expert verified
The pH of the buffer is approximately 7.02.

Step by step solution

01

Understanding Buffers

A buffer solution consists of a weak acid and its conjugate base. In this problem, \( \mathrm{KH}_2 \mathrm{PO}_4 \) acts as the weak acid (\( \mathrm{H}_2 \mathrm{PO}_4^- \)) and \( \mathrm{Na}_2 \mathrm{HPO}_4 \) provides the conjugate base (\( \mathrm{HPO}_4^{2-} \)).
02

Calculating pKa

We need the \( \mathrm{pK_a} \) of \( \mathrm{H}_2 \mathrm{PO}_4^- \) to use in the Henderson-Hasselbalch equation. The \( \mathrm{pK_a} \) of \( \mathrm{H}_2 \mathrm{PO}_4^- \) is approximately 7.2.
03

Applying the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is \( \mathrm{pH} = \mathrm{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \([\text{A}^-]\) is the concentration of \( \mathrm{HPO}_4^{2-} \) (0.10 M) and \([\text{HA}]\) is the concentration of \( \mathrm{H}_2 \mathrm{PO}_4^- \) (0.15 M).
04

Calculating the ratio

Calculate \( \log \left( \frac{0.10}{0.15} \right) \). The result is \( \log \left( \frac{2}{3} \right) \approx -0.176 \).
05

Calculating the pH

Substitute the values into the Henderson-Hasselbalch equation: \[ \mathrm{pH} = 7.2 + (-0.176) = 7.024 \].

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a key tool for calculating the pH of buffer solutions. It is derived from the acid dissociation constant expression, which describes how an acid donates protons to become its conjugate base. When a buffer consists of a weak acid and its conjugate base, this equation relates the concentration of each component to the pH. The equation is written as follows: \[\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]Where:- \(\text{pH}\) is the measure of acidity or basicity of the solution.- \(\text{pKa}\) is the negative logarithm of the acid dissociation constant of the weak acid.- \([\text{A}^-]\) signifies the concentration of the conjugate base.- \([\text{HA}]\) represents the concentration of the weak acid itself.This equation helps estimate the pH by acknowledging that a buffer can resist changes in pH when acids or bases are added. This is especially useful in biological and chemical laboratories to maintain ideal conditions for reactions.
buffer solution
A buffer solution is a special type of solution that can resist changes in pH upon adding small amounts of acids or bases. It usually consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. The presence of both forms enables the solution to adjust hydrogen ion concentrations efficiently, thus stabilizing the pH.For example, the exercise mentions a buffer consisting of \(0.10 \; M\; \text{Na}_2\text{HPO}_4 / 0.15 \; M \; \text{KH}_2\text{PO}_4\). Here, the solution maintains pH stability because the weak acid, \(\text{H}_2\text{PO}_4^-\), can donate hydrogen ions when needed, while its conjugate base, \(\text{HPO}_4^{2-}\), can accept hydrogen ions. This ability to neutralize added acids or bases without significant pH changes makes buffers crucial in many applications.
weak acid and conjugate base pair
A weak acid does not completely dissociate into its ions in water, which provides it the capability to act as a buffer. In a buffer, the weak acid is paired with its conjugate base—its deprotonated form. This pair acts to maintain the pH by equilibrium.Consider \(\text{KH}_2\text{PO}_4\) in the exercise. It acts as a weak acid, \(\text{H}_2\text{PO}_4^-\), that only partially dissociates in solution, influencing the acidity. Its conjugate base, provided by \(\text{Na}_2\text{HPO}_4\), is \(\text{HPO}_4^{2-}\). When an acidic or basic substance is added to the buffer, this pair works internally to keep hydrogen ion concentration changes minimal. This synergistic interaction allows the buffer to withstand pH shifts and is crucial for maintaining conditions when conducting scientific experiments.
pKa value
The \(\text{pKa}\) value is a critical measure of the strength of an acid. It is the negative logarithm of the acid's dissociation constant \((K_a)\), and it signifies how easily an acid donates protons. Thus, the lower the \(\text{pKa}\), the stronger the acid since it dissociates more readily.In the context of the exercise, the \(\text{pKa}\) value of \(\text{H}_2\text{PO}_4^-\) is mentioned as about 7.2. This number indicates the acid's relative strength and predictably helps determine the buffer's pH using the Henderson-Hasselbalch equation. The \(\text{pKa}\) value also informs us about the buffer's most effective pH range, typically one pH unit above and below \(\text{pKa}\). This element underscores its role in creating environments with stable hydrogen ion concentrations necessary for various chemical processes.

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Most popular questions from this chapter

Using only a pH meter, water, and a graduated cylinder, how would you distinguish between an acid solution and a buffer solution at the same \(\mathrm{pH}\) ?

The \(K_{\mathrm{a}}\) of a certain indicator is \(2.0 \times 10^{-6} .\) The color of HIn is green and that of \(\mathrm{In}^{-}\) is red. A few drops of the indicator are added to an \(\mathrm{HCl}\) solution, which is then titrated against an \(\mathrm{NaOH}\) solution. At what \(\mathrm{pH}\) will the indicator change color?

Briefly describe what happens in an acid-base titration.

Water containing \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions is called hard water and is unsuitable for some household and industrial use because these ions react with soap to form insoluble salts, or curds. One way to remove the \(\mathrm{Ca}^{2+}\) ions from hard water is by adding washing soda \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)\). (a) The molar solubility of \(\mathrm{CaCO}_{3}\) is \(9.3 \times 10^{-5} \mathrm{M}\). What is its molar solubility in a \(0.050 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) solution? (b) Why are \(\mathrm{Mg}^{2+}\) ions not removed by this procedure? (c) The \(\mathrm{Mg}^{2+}\) ions are removed as \(\mathrm{Mg}(\mathrm{OH})_{2}\) by adding slaked lime \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]\) to the water to produce a saturated solution. Calculate the \(\mathrm{pH}\) of a saturated \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution. (d) What is the concentration of \(\mathrm{Mg}^{2+}\) ions at this \(\mathrm{pH}\) ? (e) In general, which ion \(\left(\mathrm{Ca}^{2+}\right.\) or \(\mathrm{Mg}^{2+}\) ) would you remove first? Why?

The molar solubility of \(\mathrm{AgCl}\) in \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\) is \(2.5 \times 10^{-8} M\). In deriving \(K_{\mathrm{sp}}\) from these data, which of the following assumptions are reasonable? (a) \(K_{\mathrm{sp}}\) is the same as solubility. (b) \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}\) is the same in \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\) as in pure water. (c) Solubility of \(\mathrm{AgCl}\) is independent of the concentration of \(\mathrm{AgNO}_{3}\). (d) \(\left[\mathrm{Ag}^{+}\right]\) in solution does not change significantly upon the addition of \(\mathrm{AgCl}\) to \(6.5 \times 10^{-3} \mathrm{M} \mathrm{AgNO}_{3}\). (e) \(\left[\mathrm{Ag}^{+}\right]\) in solution after the addition of \(\mathrm{AgCl}\) to \(6.5 \times 10^{-3} \mathrm{M}\) \(\mathrm{AgNO}_{3}\) is the same as it would be in pure water.
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