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Determine the [OH-], [H+], pH, and pOH of each of the following: a. 0.29 M Sr(OH)2...

Determine the [OH-], [H+], pH, and pOH of each of the following: a. 0.29 M Sr(OH)2 b. A 2.50 L solution prepared by dissolving 15.6 g of KOH in distilled water. c. A 250 ml solution prepared by dissolving 42.1 g of NaOH in distilled water d. A 300 ml solution prepared by dissolving 5.00 g of HI in 300ml of a 0.796 M solution of HNO3?

Solutions

Expert Solution

a. 0.29 M Sr(OH)2

Sr(OH)2 = Sr2+ 2 [OH-]

Thus 0.29 M Sr(OH)2 = 2* 0.29 M [OH-]

= 0.58 M [OH-]

pOH = - log [OH-]

= 0.237

pH + pOH = 14

pH = 14-0.237

= 13.76

[H+] = 10^-pH

=10^- 13.76

= 1.74*10^-14

b. A 2.50 L solution prepared by dissolving 15.6 g of KOH in distilled water

moles of KOH = 15.6 g / molar mass; 56.1056 g/mol

= 0.278 moles

Molarity = number of moles / volume in L

= 0.278 moles / 2..50 L

= 0.11 M

KOH = K+ + OH-

[OH-]= 0.11 M

pOH = - log [OH-]

= 0.95

pH + pOH = 14

pH = 14-0.95

= 13.05

[H+] = 10^-pH

=10^- 13.05

=8.91*10^-14

c. A 250 ml solution prepared by dissolving 42.1 g of NaOH in distilled water

moles of NaOH = 42.1 g / molar mass; 39.997 g/mol

= 1.053 moles

Molarity = number of moles / volume in L

= 1.053 moles / 0.250L

= 4.2 M

NaOH = Na+ + OH-

[OH-]= 4.2 M

pOH = - log [OH-]

= 0.62

pH + pOH = 14

pH = 14-0.62

= 13.38

[H+] = 10^-pH

=10^- 13.38

=4.17*10^-14

d. A 300 ml solution prepared by dissolving 5.00 g of HI in 300ml of a 0.796 M solution of HNO3?

Moles of HI= 5.00 g/ 127.911 g/ mole

= 0.039 moles HI

Or 0.039 H+

Moles of HNO3 = Molarity * volume in L

= 0.796 *0.300

= 0.2388 HNO3

0.2388 H+

Total H+ = 0.2778 moles

Molarity = 0.2778 /0.300 L

= 0.926 M

pH = -log [H+]

= 0.033

pH + pOH = 14

pOH = 14-0.033

= 13.967

[OH-] = 10^-pOH

=10^- 13.967

=1.08*10^-14

queen_honey_blossom answered 10 months ago

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