/*! 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 159 Is an aqueous solution of \(\mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 159

Is an aqueous solution of \(\mathrm{NaHSO}_{4}\) acidic, basic, or neutral? What reaction occurs with water? Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of \(\mathrm{NaHSO}_{4}\).

Short Answer

Expert verified
The aqueous solution of NaHSO鈧� is acidic due to the HSO鈧勨伝 ions donating a proton to water molecules, forming H鈧僌鈦� ions. The pH of a 0.10 M NaHSO鈧� solution is approximately 6.5.

Step by step solution

01

Write the dissociation reaction of NaHSO鈧� in water

NaHSO鈧� is a salt formed from the reaction of a strong base (NaOH) and a weak acid (H鈧係O鈧�). Let's write the dissociation reaction of NaHSO鈧� in water: NaHSO鈧� (aq) 鈬� Na鈦� (aq) + HSO鈧勨伝 (aq)
02

Determine the reaction of HSO鈧勨伝 ions with water

As NaHSO鈧� dissociates in water, we should consider the possible reaction of HSO鈧勨伝 ions with water molecules. If HSO鈧勨伝 donates a proton (H鈦�) to water, the solution will be acidic. If HSO鈧勨伝 accepts a proton from water, the solution will be basic. Since HSO鈧勨伝 is the conjugate base of a weak acid (H鈧係O鈧�), it will tend to donate a proton to water: HSO鈧勨伝 (aq) + H鈧侽 (l) 鈬� H鈧僌鈦� (aq) + SO鈧劼测伝 (aq)
03

Write the equilibrium expression and calculate the pH

Now, we need to write the expression for the equilibrium constant, K鈧�, for the reaction of HSO鈧勨伝 with water and use it to calculate the pH of the 0.10 M NaHSO鈧� solution. First, we need to find the value of K鈧� for the reaction. The K鈧� of H鈧係O鈧� is 1.0 x 10虏, and since K鈧怟鈧愨倎 is approximately equal to K鈧�(Kw), we can find K鈧愨倎 for HSO鈧勨伝: K鈧愨倎 = Kw / K鈧� = 1.0 x 10鈦宦光倓 / 1 x 10虏 = 1.0 x 10鈦烩倎鈧� Now, we can write the equilibrium expression for the reaction: K鈧愨倎 = [H鈧僌鈦篯[SO鈧劼测伝] / [HSO鈧勨伝] Since the concentration of NaHSO鈧� is 0.10 M and it dissociates completely in water, the initial concentration of HSO鈧勨伝 ions can be assumed to be 0.10 M. Let x represent the equilibrium concentration of H鈧僌鈦�: 1.0 x 10鈦烩倎鈧� = x虏 / (0.10 - x) Assuming x << 0.10, we can simplify the expression: x虏 鈮� 1.0 x 10鈦烩倎鲁 x 鈮� 鈭�(1.0 x 10鈦烩倎鲁) = 1.0 x 10鈦烩伓.鈦� Since x represents the concentration of H鈧僌鈦�, the pH can be calculated as: pH = -log([H鈧僌鈦篯) = -log(1.0 x 10鈦烩伓.鈦�) 鈮� 6.5
04

Conclusion

The aqueous solution of NaHSO鈧� is acidic, as the HSO鈧勨伝 ions donate a proton to water molecules, forming H鈧僌鈦� ions. The pH of a 0.10 M NaHSO鈧� solution is approximately 6.5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Dissociation of NaHSO4
When sodium hydrogen sulfate (NaHSO4) dissolves in water, it undergoes a process called dissociation. This is where NaHSO4 breaks down into its constituent ions. The equation for this dissociation is:

NaHSO4 (aq) NaH 鈬� Na^+ (aq) + HSO4^- (aq).

The dissociation of NaHSO4 is essential in determining the nature of its solution because it releases hydrogen sulfate ions (HSO4^-), which in turn can affect the solution's pH. As NaHSO4 comes from the neutralization of a strong base and a weak acid, the resulting solution can be unambiguously understood as having a pH that is not neutral.

Understanding the behavior of the HSO4^- ion in water is crucial, as its ability to donate a proton is what ultimately leads to the acidic nature of the NaHSO4 solution.
Acidic and Basic Solutions
Determining whether a solution is acidic, basic, or neutral is based upon the concentration of hydronium ions (H3O^+) relative to hydroxide ions (OH^-). Acidic solutions have more hydronium ions than hydroxide ions, while basic solutions have more hydroxide ions.

In the case of NaHSO4, the dissociated hydrogen sulfate ions (HSO4^-) can donate protons to water, forming hydronium ions (H3O^+). This proton-donating reaction can be represented as:

HSO4^- (aq) + H2O (l) 鈬� H3O^+ (aq) + SO4^2- (aq).

This shows that the NaHSO4 solution is indeed acidic due to the formation of H3O^+ ions. So, when a solution of NaHSO4 is mixed with water, inspecting the behavior of HSO4^- reveals a clear tendency for the solution to be more acidic.
pH Calculation
The pH of a solution is a numeric scale used to specify the acidity or basicity of an aqueous solution. It is defined as the negative logarithm of the concentration of hydronium ions (H3O^+).

For the NaHSO4 solution, the pH can be calculated using the expression:
pH = -log([H3O^+]).

By using the equilibrium constant (Ka1), we can estimate the concentration of those hydronium ions in the solution and subsequently determine the pH. The provided steps in the textbook solution show how to derive the concentration of H3O^+ from the equilibrium expression, which greatly simplifies to X When x is smaller than the initial concentration of HSO4^- (0.10 M).

Following the simplification, we can calculate the pH, as shown:

pH [ x H3O^+nly consider the significant figures and rationale behind the approximation of X in the final answer. This detailed look into the pH calculation provides students with a deeper understanding of the concept and its application in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One mole of a weak acid HA was dissolved in \(2.0 \mathrm{~L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be \(0.45 M .\) Calculate \(K_{\mathrm{a}}\) for HA.

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\), a triprotic acid.

Identify the Lewis acid and the Lewis base in each of the following reactions. a. \(\mathrm{B}(\mathrm{OH})_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}^{+}(a q)\) b. \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)\) c. \(\mathrm{BF}_{3}(g)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BF}_{4}^{-}(a q)\)

Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. \(\mathrm{HIO}_{3}, \mathrm{HBrO}_{3}\) c. HOCl, HOI b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\) d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10 \mathrm{M}\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.