Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is theof a saturated solution of Ca(OH)2?

Short Answer

Expert verified

The pH of the solution is 12.14

Step by step solution

01

Define saturated solution

This state is when the solution has reached a point in which no more solute can be added. The addition of solute after this point would result in a solid precipitate or gas being released. Such a mixture is called a saturated solution

02

The reaction of the saturated solution

Reaction of

Ca(OH)2(s) ⇌ Ca2+ (aq) +2OH- (aq)

Reaction of

Kspof Ca(OH)2is Ksp = [Ca2+] . [OH-] = 1.3.10-6

The molar solubility of Ca2+ is ‘s’

The molar solubility of OH-is ‘2s’

03

  Step 3: The pH-saturated solution Ca(OH)2

Let us find the value of

\begin{aligned}{{K_{sp}}=\left[{C{a^{2+}}}\right]\cdot{{\left[{O{H^-}}\right]}^2}}\\{1.3\cdot{{10}^{-6}}=s\cdot{{(2s)}^2}}\\{4{s^3}=1.3\cdot{{10}^{-6}}}\\{s=6.875\cdot{{10}^{-3}}}\\{\left[{O{H^-}}\right]=2s=1.375\cdot{{10}^{-2}}{\rm{M}}}\end{aligned}

04

Finding the pH

Let us calculate

\begin{aligned}{pOH=-\log\left[{O{H^-}}\right]}\\{=-\log\left[{1.375\cdot{{10}^{-2}}}\right]}\\{=1.86}\end{aligned}

pH Of the saturated solution

\begin{aligned}{pH=14-pOH}\\{=14-1.86}\\{=12.14}\end{aligned}

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 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

Question: How many grams of Milk of Magnesia, \(Mg{(OH)_2}(s)(58.3g/mol)\)would be soluble in 200 mL of water. \({K_{sp}}\)=\(7.1 \times 1{0^{ - 12}}\) Include the ionic reaction and the expression for \({K_{sp}}\)in your answer\(\left( {{K_w} = 1 \times 1{0^{ - 14}} = } \right.\)\(\left. {\left( {{H_3}{O^ + }} \right)\left( {O{H^ - }} \right)} \right)\)

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

\(\begin{array}{l}(a)BaSi{F_6},0.026\;g/100\;mL(contains Si{F_6}^2 - ions)\\(b)Ce{\left( {I{O_3}} \right)_4},1.5 \times 1{0^{ - 2}}\;g/100\;mL\\(c)G{d_2}{\left( {S{O_4}} \right)_3},3.98\;g/100\;mL\\(d){\left( {N{H_4}} \right)_2}PtB{r_6},0.59\;g/100\;mL(contains PtB{r_6}^{2 - } ions)\end{array}\)

A volume of \(50mL\) of \(1.8MN{H_3}\)is mixed with an equal volume of a solution containing\(0.95g\;of\;MgC{l_2}\). What mass of \(N{H_4}Cl\)must be added to the resulting solution to prevent the precipitation of \(Mg{(OH)_2}?\)

Question: Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from seawater by the following sequence of reactions:

\(\begin{array}{*{20}{c}}{M{g^{2 + }}(aq) + Ca{{(OH)}_2}(aq) \to Mg{{(OH)}_2}(s) + C{a^{2 + }}(aq)}\\{Mg{{(OH)}_2}(s) + 2HCl(aq) \to MgC{l_2}(s) + 2{H_2}O(l)}\end{array}\)

\(MgC{l_2}(l)\mathop \to \limits^{\;electrolysis\;} Mg(s) + C{l_2}(g)\)

Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium \(M{g^{2 + }}(aq)\)by mass. What mass, in kilograms, \(Ca{(OH)_2}\)is required to precipitate 99.9% of the magnesium in 1.00 × 103 L of seawater?

Question: A solution contains \(1.0 \times 1{0^{ - 5}}\)mol of KBr and 0.10 mol of KCl per liter. \(AgN{O_3}\)is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?

See all solutions

Recommended explanations on Chemistry Textbooks

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