Part A: A solution has [OH?] = 1.3×10?2 M . Use the ion product constant of wate

Question

Part A:

A solution has [OH?] = 1.3×10?2 M . Use the ion product constant of water, Kw=[H3O+][OH?]   to find the [H3O+] of the solution. Express your answer to two significant figures.

Part B:

In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 10.75 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH , note the corresponding OH? ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L )" above the graphic.

pOH=?log[OH?]

Find the pOH of the solution.

Part c:

https: 'session masteringchemistry.corm m ct t Chegg Study | Guided Solution... MasteringChemistry: Chapt. Ra nk the solutions by increasin... MasteringChemistry: Course H. File Edit View Favorites Tools Help 23 old Fashioned Coconut. Q Biotest #4 flashcards Qui Q Search chapter 8 test che.. Chemistry is EASY!Howd. Chem 100 Fall 2015 Chapter 14 Homework Core Chemistry Skill: Writing Equations for Reactions of Acids and Bases QChemistry Mid-Term Mul Page. Safety. Tools E- Help | Close Resources « previous 15 of 22 next» rag the appropriate reactions to their respective bins MgCl2 (aq) + H2O (1) AgCl (s) + NaNO3 (aq) PbI2 (s+ KNO3 (aq) Mg(OH)2 (aq) + HCl (aq) CH,COOH (aq) + KOH (aq)-> CH3COOK (aq) + H2O (1) AgNO3 (aq) + NaCl (aq)-> HBr (aq) +NaOH (aq)NaBr (a) H2O ) Pb(NO3)2 (aq) + KI (aq)- Bacl2 (aq) + Na,SO4 (aq)--> BaSO4 (s) + NaCl (aq) > Neutralization reaction Precipitation reaction reset help 10096 4:11 PM 12/5/2015

Explanation / Answer

A)Given [OH-] = 1.3x 10-2M and we have ionic product of water Kw = 1.0 x 10-14

Hence [H3O+] = 1.0x 10-14/ 1.3 x 10-2 = 7.69x 10-13 M

B) Given pH = 10.75 thus pOH = 14 - pH = 3.25

Alternatively [H3O+] in the solution = antilog [-10.75] = 1.77x 10 -11 M and then [OH-] form ionic product is 5.65 x 10-4M

Thus pOH = - log [5.65 x 10-4] = 3.25

C)    Neutralization reactions are the reactions in which an acid reacts with a base to give an aqueous salt and water usually without any solid precipitation.

   Mg(OH)2 + 2HCl -----> MgCl2 + 2H2O

CH3COOH + KOH --------> CH3COOK + H2O

HBr + NaOH ---------> NaBr + H2O

Precipitation reactions are the ones in which two aqueous salts react to form onesalt as solid precipitate and another aqueous salt .[double decomposition]

AgNO3 +NaCl ------> AgCl(s) + NaNO3

Pb(NO3)2 + 2KI --------> PbI2(s) +2KNO3

BaCl2 + Na2SO4 --------> BaSO4 + 2NaCl

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