Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 45

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2}\) (b) 2.250 \(\mathrm{g}\) of LiOH in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 1.00\) mL of 0.175 M NaOH diluted to \(2.00 \mathrm{L},\) (d) a solution formed by adding 5.00 \(\mathrm{mL}\) of 0.105 \(\mathrm{M} \mathrm{KOH}\) to 15.0 \(\mathrm{mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}.\)

Short Answer

Expert verified
In summary, the solutions for the given exercise are: (a) \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{pH} = 11.48\) (b) \(\left[\mathrm{OH}^{-}\right] = 0.376 \mathrm{M}\) and \(\mathrm{pH} = 13.576\) (c) \(\left[\mathrm{OH}^{-}\right] = 8.75\times10^{-5} \mathrm{M}\) and \(\mathrm{pH} = 9.942\) (d) \(\left[\mathrm{OH}^{-}\right] = 0.158\mathrm{M}\) and \(\mathrm{pH} = 13.199\)

Step by step solution

01

Determine the number of moles of OH- ions

Each mole of Sr(OH)\(_{2}\) will dissociate into 2 moles of OH- ions. Therefore, multiply the concentration by two to find the concentration of hydroxide ions: \(\left[\mathrm{OH}^{-}\right]\) = \(2(1.5 \times 10^{-3}) = 3.0 \times 10^{-3} \text{M}\)
02

Calculate the pOH

Use the equation \(\mathrm{pOH} = - \log_{10}[\mathrm{OH}^{-}]\) to calculate the pOH: \(\mathrm{pOH} = - \log_{10}(3.0 \times 10^{-3}) = 2.52\)
03

Calculate the pH

Lastly, use the equation \(\mathrm{pH} = 14 - \mathrm{pOH}\) to find the pH: \(\mathrm{pH} = 14 - 2.52 = 11.48\) Part (a) Results: \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{pH} = 11.48\) Part (b): 2.250 g of LiOH in 250.0 mL of solution
04

Calculate mole of LiOH

We first need to convert the mass of LiOH to moles of LiOH using its molar mass (Li: 6.94 g/mol, O: 16.00 g/mol, H: 1.01 g/mol): Moles of LiOH = \(\frac{2.250\,\text{g}}{6.94\,\text{g/mol} + 16.00\,\text{g/mol} + 1.01\,\text{g/mol}}=\frac{2.250\,\text{g}}{23.95\,\text{g/mol}}\approx 0.0940\,\text{moles}\)
05

Determine the moles of OH- ions

LiOH dissociates into Li+ and OH- ions in a 1:1 ratio. Therefore, the moles of OH- ions will be equal to the moles of LiOH.
06

Calculate the concentration of OH- ions

Divide the moles of OH- ions by the volume of the solution (in Liters) to find the concentration of hydroxide ions: \(\left[\mathrm{OH}^{-}\right] = \frac{0.0940\,\text{moles}}{0.250\,\text{L}} = 0.376\,\text{M}\)
07

Calculate the pOH and pH

Using the same equations as before, we find the pOH: \(\mathrm{pOH} = - \log_{10}(0.376) \approx 0.424\) And then find the pH: \(\mathrm{pH} = 14 - 0.424 \approx 13.576\) Part (b) Results: \(\left[\mathrm{OH}^{-}\right] = 0.376 \mathrm{M}\) and \(\mathrm{pH} = 13.576\) Part (c): 1.00 mL of 0.175 M NaOH diluted to 2.00 L
08

Calculate moles of OH- ions

Calculate the moles of OH- ions from the initial concentration: Moles of OH- ions = (0.175 mol/L)(0.001 L) = 1.75 x 10^(-4) mol
09

Calculate the concentration of OH- ions

Divide the moles of OH- ions by the total volume (in Liters) to find the hydroxide ion concentration: \(\left[\mathrm{OH}^{-}\right] = \frac{1.75\times10^{-4}\,\text{moles}}{2.00\,\text{L}} = 8.75\times10^{-5}\,\text{M}\)
10

Calculate the pOH and pH

Using the equations as before, find the pOH: \(\mathrm{pOH} = - \log_{10}(8.75\times10^{-5}) \approx 4.058\) And then find the pH: \(\mathrm{pH} = 14 - 4.058 = 9.942\) Part (c) Results: \(\left[\mathrm{OH}^{-}\right] = 8.75\times10^{-5} \mathrm{M}\) and \(\mathrm{pH} = 9.942\) Part (d): a solution formed by adding 5.00 ml of 0.105 M KOH to 15.0 ml of \(9.5 \times 10^{-2} \mathrm{M Ca(OH)_{2}}\)
11

Calculate moles of OH- ions in each solution

Calculate the total moles of OH- ions from both the KOH and Ca(OH)\(_{2}\) solutions: Moles of OH- ions from KOH = (0.105 mol/L)(0.005 L) = 5.25 x 10^(-4) moles Moles of OH- ions from Ca(OH)\(_{2}\) = 2(9.5 x 10^(-2) mol/L)(0.015 L) = 2.85 x 10^(-3) moles
12

Calculate the total volume

Add together the volumes of both solutions: Total Volume = 5.00 mL + 15.0 mL = 20.0 mL = 0.020 L
13

Calculate the concentration of OH- ions

Add the moles of OH- ions from both solutions, and then divide by the total volume to find the hydroxide ion concentration: \(\left[\mathrm{OH}^{-}\right] = \frac{(5.25\times10^{-4}\,\text{moles} + 2.85\times10^{-3}\,\text{moles})}{0.020\,\text{L}} \approx 0.158\,\mathrm{M}\)
14

Calculate the pOH and pH

Using the equations as before, find the pOH: \(\mathrm{pOH} = - \log_{10}(0.158) \approx 0.801\) And then find the pH: \(\mathrm{pH} = 14 - 0.801 = 13.199\) Part (d) Results: \(\left[\mathrm{OH}^{-}\right] = 0.158\mathrm{M}\) and \(\mathrm{pH} = 13.199\) In conclusion, the solutions for the given exercise are as follows: (a) \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{pH} = 11.48\) (b) \(\left[\mathrm{OH}^{-}\right] = 0.376 \mathrm{M}\) and \(\mathrm{pH} = 13.576\) (c) \(\left[\mathrm{OH}^{-}\right] = 8.75\times10^{-5} \mathrm{M}\) and \(\mathrm{pH} = 9.942\) (d) \(\left[\mathrm{OH}^{-}\right] = 0.158\mathrm{M}\) and \(\mathrm{pH} = 13.199\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) \(\operatorname{Fe}\left(\mathrm{ClO}_{4}\right)_{3}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(a q)+3 \mathrm{ClO}_{4}^{-}(a q)\) (b) \(\mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\) (c)\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(g)+\mathrm{BF}_{3}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NBF}_{3}(s)\) (d)\(\operatorname{HIO}(l q)+\mathrm{NH}_{2}^{-}(l q) \Longrightarrow \mathrm{NH}_{3}(l q)+\mathrm{IO}^{-}(l q)\) (\(lq\) denotes liquid ammonia as solvent)

Consider two solutions, solution \(\mathrm{A}\) and solution \(\mathrm{B} .\left[\mathrm{H}^{+}\right]\) in solution \(\mathrm{A}\) is 250 times greater than that in solution B. What is the difference in the pH values of the two solutions?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: \((\mathbf{a}) 0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{g}\) of \(\mathrm{KOH}\) in 500.0 mL of solution, ( c ) 10.0 \(\mathrm{mL}\) of 0.0105 \(\mathrm{MCa}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{mL},(\mathbf{d})\) a solution formed by mixing 20.0 \(\mathrm{mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with 40.0 \(\mathrm{mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}.\)

A particular sample of vinegar has a pH of \(2.90 .\) If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\) calculate the concentration of acetic acid in the vinegar.

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{Br}^{-} .\) An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation atequilibrium, in units of molarity?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

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.

Sign-up for free