/*! 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 32 What is the pH at \(25^{\circ} \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 32

What is the pH at \(25^{\circ} \mathrm{C}\) of a solution that results from mixing equal volumes of a \(0.05 M\) solution of acetic acid and a \(0.025 M\) solution of sodium hydroxide?

Short Answer

Expert verified
Answer: The pH of the mixed solution is approximately 4.74.

Step by step solution

01

Write down the balanced chemical equation

The first step is to write down the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH). CH3COOH + NaOH -> CH3COONa + H2O The equation shows that acetic acid reacts with sodium hydroxide to form sodium acetate (CH3COONa) and water.
02

Calculate the moles of reactants

To find the concentration of ions after the reaction, we need to calculate the moles of the reactants (acetic acid and sodium hydroxide) in the solution. As equal volumes of the two solutions are mixed, let's assume the volume of each solution is V. Moles of acetic acid = M (concentration) x V (volume) = 0.05 M x V = 0.05V Moles of sodium hydroxide = M (concentration) x V (volume) = 0.025 M x V = 0.025V
03

Determine the limiting reactant and calculate moles of products

From the balanced equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide. However, in our case, the moles of acetic acid are more than the moles of sodium hydroxide, so sodium hydroxide is the limiting reactant. The moles of acetic acid and sodium hydroxide react in a 1:1 ratio, so the moles of sodium acetate produced is 0.025V. The moles of remaining acetic acid is 0.05V - 0.025V = 0.025V.
04

Calculate concentrations of products and remaining reactant

Now, the total volume of the solution after mixing is 2V. To find the concentration of species in the solution, we can use the formula: Concentration = Moles/V, where V is the total volume of the solution. Concentration of sodium acetate (CH3COONa) = 0.025V / 2V = 0.0125 M Concentration of acetic acid (CH3COOH) = 0.025V / 2V = 0.0125 M
05

Calculate the pH of the mixed solution

We can use the equation for acetic acid ionization to help us find pH. CH3COOH (aq) <-> CH3COO- + H+ Since acetic acid is a weak acid, we can use the Ka expression: Ka = (molar concentration of CH3COO-)(molar concentration of H+)/(molar concentration of CH3COOH) For acetic acid, Ka = 1.8 x 10^-5 The concentration of CH3COO- and CH3COOH is the same in the solution, let's denote it as x: 1.8 x 10^-5 = (x)(H+)/((0.0125-x)) As x is very small in comparison to 0.0125, we can simplify the equation as: 1.8 x 10^-5 = (x)(H+)/0.0125 To find H+ concentration, we can solve for H+: H+ = (1.8 x 10^-5)(0.0125)/x Finally, to find the pH, we use the equation pH = -log(H+): pH = -log((1.8 x 10^-5)(0.0125)/x) The pH of the mixed solution is approximately 4.74.

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!

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

Exactly \(1.00 \mathrm{g}\) of \(\mathrm{FeCl}_{3}\) is dissolved in each of four \(0.500 \mathrm{L}\) samples: \(1 M \mathrm{HNO}_{3}, 1 \mathrm{MHNO}_{2}, 1 \mathrm{MCH}_{3} \mathrm{COOH},\) and pure water. Is the concentration of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) ions the same in all four solutions? Explain why or why not.

For each set of three acids, which one should you use to make a buffer with the given pH? a. Select from acetic acid \(\left(\mathrm{p} K_{2} 4.75\right),\) fluoroacetic acid \(\left(\mathrm{p} K_{\mathrm{a}} 2.59\right)\) and hypochlorous acid \(\left(\mathrm{p} K_{\mathrm{a}} 7.54\right)\) to make a buffer with a pH of 5.2. b. Select from formic acid \(\left(\mathrm{p} K_{2}=3.75\right),\) hypobromous acid \(\left(\mathrm{p} K_{\mathrm{a}}=7.54\right),\) and boric acid \(\left(\mathrm{p} K_{2} 9.27\right)\) to make a buffer with a pH of 8.0.

For each titration, predict whether the pH of the equivalence point is less than, equal to, or greater than 7 a. Quinine titrated with nitric acid b. Pyruvic acid titrated with calcium hydroxide c. Hydrobromic acid titrated with strontium hydroxide

Which has a higher \(\mathrm{pH}\), a saturated solution of magnesium hydroxide \(\left(K_{\mathrm{sp}}=5.6 \times 10^{-12}\right)\) or a saturated solution of calcium hydroxide \(\left(K_{\mathrm{sp}}=4.7 \times 10^{-6}\right) ?\)

A \(25.0 \mathrm{mL}\) sample of \(0.100 M\) acetic acid is titrated with \(0.125 \mathrm{M} \mathrm{NaOH}\) at \(25^{\circ} \mathrm{C} .\) What is the \(\mathrm{pH}\) of the solution after \(10.0,20.0,\) and \(30.0 \mathrm{mL}\) of the base have been added?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

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.