Part C: At a certain temperate, the pH of a neutral solution is 7.26. What is th

Question

Part C: At a certain temperate, the pH of a neutral solution is 7.26. What is the value of Kw at that temperate?

t Acid-Base Relationships in Water Water ionizes by the equation (l) OH (aq) The extent of the reaction is small in pure water and dilute aqueous solutions. This reaction creates the following relationship between H+ and OH Keep in mind that, like all equilibrium constants, the value of Kw changes with temperature what is the H concentration for an aqueous solution with poH 3.76 at 25 C? Express your answer to two significant figures and include the appropriate units. Value Units Submit Hints My Answers Give Up Review Part Part B Arrange the following aqueous solutions, all at 25 C n order of decreasing acidity. Rank from most acidic to most basic. To rank items as equivalent, overlap them. Reset Help poH 8.55 0.00 23 M HCl pH 545 0.0018 M KOH Most acidic Most basic

Explanation / Answer

POH   = 3.76

PH = 14-POH

      = 14-3.76

PH    = 10.24

-log[H+] = 10.24

[H+]       = 10^-10.24   = 5.75*10^-11 M

Part-B

POH = 8.55

PH   = 14-POH

      = 14-8.55 = 5.45

HCl ------------> H+ + Cl-

0.0023M           0.0023M

[H+]     = [HCl]

[H+]    = 0.0023M

PH   = -log[H+]

       = -log0.0023   = 2.6382

PH = 5.45

KOH ----------> K+ + OH-

0.0018M                  0.0018M

[OH-]    = [KOH]

[OH-]   = 0.0018M

POH = -log0.0018   = 2.7447

PH   = 14-POH

       = 14-2.7447   = 11.2553

0.0023M HCl more acidic

0.0023M HCl >POH =8.55 ,PH= 5.45 >0.0018M KOH is more basic

part-C

   PH   = 7.26

-log[H+] = 7.26

     [H+]   = 10^-7.26    = 5.5*10^-8M

in neutral solution [H+] = [OH-]

                            [OH-] = 5.5*10^-8M

Kw   = [H+][OH-]

         = 5.5*10^-8 * 5.5*10^-8   = 3*10^-15 M2

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