Question

In: Chemistry

Calculate the [H3O+] of each aqueous solution with the following [OH−] . Express your answer using...

Calculate the [H3O+] of each aqueous solution with the following [OH−] . Express your answer using two significant figures.

1.Vinegar, 1.2 x 10^-11M, Express your answer using two significant figures.

2. Urine, 2.3x10^-9 M, Express your answer using two significant figures.

3. Ammonia, 5.8×10⁻³ M, Express your answer using two significant figures.

4. NaOH, 4.0×10⁻² M, Express your answer using two significant figures.

Solutions

Expert Solution

we have below equation to be used:

[H3O+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H3O+] = (1.0*10^-14)/[OH-]

[H3O+] = (1.0*10^-14)/1.2E-11

[H3O+] = 8.3*10^-4 M

Answer: 8.3*10^-4 M

we have below equation to be used:

[H3O+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H3O+] = (1.0*10^-14)/[OH-]

[H3O+] = (1.0*10^-14)/2.3E-9

[H3O+] = 4.3*10^-6 M

Answer: 4.3*10^-6 M

we have below equation to be used:

[H3O+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H3O+] = (1.0*10^-14)/[OH-]

[H3O+] = (1.0*10^-14)/0.0058

[H3O+] = 1.7*10^-12 M

Answer: 1.7*10^-12 M

we have below equation to be used:

[H3O+] = Kw/[OH-]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[H3O+] = (1.0*10^-14)/[OH-]

[H3O+] = (1.0*10^-14)/0.04

[H3O+] = 2.5*10^-13 M

Answer: 2.5*10^-13 M

queen_honey_blossom answered 1 year ago

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