Calculate [OH^-] for solution 2.97times10^-3 m Sr(OH)_2. Express your answer to

Question

Calculate [OH^-] for solution 2.97times10^-3 m Sr(OH)_2. Express your answer to three significant figures and include the appropriate units. Calculate [H_3O^+] for solution 2.97times10^-3 M Sr(OH)_2. Express your answer to two significant figures and include the appropriate units. Calculate [H_3O^+] for solution 7.6times10^-4 M HI. Express your answer to two significant figures and include the appropriate units. Calculate [OH^-] for solution 7.6times10^-4 M HI. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Part E

[Sr(OH)2] = 2.97 x 10-3 M

Sr(OH)2 is a strong base, and therefore it will dissociate completely. Therefore, the hydroxide ion concentration in moles per liter, [OH-], will be twice the concentration of the base.

[OH-] = 2 x (2.97 x 10-3 M) = 5.9 x 10-3 M

Part F

[Sr(OH)2] = 2.97 x 10-3 M

[OH-] = 5.94 x 10-3 M (from answer of Part E)

pOH = - log[OH-] = - log (5.94 x 10-3) = 2.23

pH + pOH = 14

pH = 14 - pOH = 14 - 2.23 = 11.77

Now,

-log [H3O+] = pH = 11.77

-[H3O+] =10-11.77 = 1.7 x 10-12

Part G

[HI] = 7.6 x 10-4M

HI is a strong base, and therefore it will dissociate completely. Therefore, the [H3O+] concentration in moles per liter, will be the concentration of the acid.

[H3O+] = 7.6 x 10-4M

Part H

[HI] = 7.6 x 10-4M

[H3O+] = 7.6 x 10-4M

Since,

[H3O+] [OH-]= 10-14

[OH-] = 10-14 / [H3O+]

= (10-14) / ( 7.6 x 10-4)

= 1.3 x 10-11 M

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