(7.5 points) The solutions in beakers X and Y were combined in beaker Z, as show

Question

(7.5 points) The solutions in beakers X and Y were combined in beaker Z, as shown in the cartoons below. Use a calculator to complete parts (b) through (d) for each solution. The volumes of the solutions in the cartoons are approximate. 1. + = 5.0 × 10-3 M : Equations [H+] × [OH1-1 × 10-14 PHloglH'] 1 Solution X (b) [OH] (c) pH = (d) circle one: acidic basic HI -3.5 x 101M approximately neutral Solution Y (a) H1 3.5x 10-11 M (b) IOH] (c) pH (d) circle one: acidic basic approximately neutral Solutionz (b) (OH] :(c) pH (d) circle one @ ch PrtScn End F8

Explanation / Answer

1) for solution X

use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(5*10^-3)
[OH-] = 2*10^-12 M

use:
pH = -log [H+]
= -log (5*10^-3)
= 2.301

use:
pOH = -log [OH-]
= -log (2*10^-12)
= 11.699

Since pH < pOH, this is acidic in nature

Answers:
[H+] = 5*10^-3
[OH-] = 2*10^-12
pH = 2.301
It is acidic

2) for solution Y

use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(3.5*10^-11)
[OH-] = 2.857*10^-4 M

use:
pH = -log [H+]
= -log (3.5*10^-11)
= 10.4559

use:
pOH = -log [OH-]
= -log (2.857*10^-4)
= 3.5441

Since pH > pOH, this is basic in nature

Answers:
[H+] = 3.5*10^-11
[OH-] = 2.857*10^-4
pH = 10.4559
It is basic

3) for Solution Z

use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(5*10^-8)
[OH-] = 2*10^-7 M

use:
pH = -log [H+]
= -log (5*10^-8)
= 7.301

use:
pOH = -log [OH-]
= -log (2*10^-7)
= 6.699

Since pH > pOH, this is basic in nature

Answers:
[H+] = 5*10^-8
[OH-] = 2*10^-7
pH = 7.301
It is basic

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