Question

In: Chemistry

Calculate the [H3O+] of each aqueous solution with the following [OH-]. 1) NaOH, 1.6x10^2 M. 2)...

Calculate the [H3O+] of each aqueous solution with the following [OH-]. 1) NaOH, 1.6x10^2 M. 2) milk of magnesia, 1.0x10^-5 M. 3) aspirin, 2.2x10^-11 M. 4) seawater, 2.5x10^-6 M.

Solutions

Expert Solution

1) NaOH (aq) -----> Na+ (aq) + OH- (aq)

2 H2O (l) -----> H3O+ (aq) + OH- (aq)

[OH-] = [NaOH] = 1.6*102 M (I think this should be 1.6*10-2 M, otherwise we shall get a negative pH which is clearly impossible).

pOH = -log [OH-] = -log (1.6*10-2) = 1.79588; pH = 14 – pOH = 14 – 1.79588 = 12.20412 ≈ 12.204

Now pH = -log [H3O+]

====> [H3O+] = 10^(-pH) = 10^(-12.204) = 6.2517*10-13 ≈ 6.25*10-13 M (ans).

2) Milk of magnesia is Mg(OH)2. Consider the dissociation:

Mg(OH)2 (aq) -------> Mg2+ (aq) + 2 OH- (aq)

2 H2O (l) -----> H3O+ (aq) + OH- (aq)

[Mg(OH)2] = 1.0*10-5 M

Therefore, [OH-] = 2*[Mg(OH)2] (follow the dissociation equation above; one mole Mg(OH)2 furnishes 2 moles of OH-). Thus,

[OH-] = 2*(1.0*10-5) = 2.0*10-5 and pOH = -log [OH-] = -log (2.0*10-5) = 4.6989

pH = 14 – pOH = 14 – 4.6989 = 9.3011 ≈ 9.30

Again, pH = -log [H3O+]

====> [H3O+] = 10^(-pH) = 10^(-9.30) = 5.012*10-10 ≈ 5.01*10-10 (ans)

3) Aspirin is acetylsalicylic acid (let us represent the same as HA) with acid dissociation constant Ka = 3.0*10-4. Write down the dissociation equation:

HA (aq) + H2O (l) ------> H3O+ (aq) + A- (aq)

2 H2O (l) ------> H3O+ (aq) + OH- (aq)

Given [OH-] = 2.2*10-11 M, pOH = -log [OH-] = - log (2.2*10-11) = 10.6575.

We know that pH + pOH = 14; therefore, pH = 14 - pOH = 14 – 10.6575 = 3.3425

Again, pH = -log [H3O+]

====> [H3O+] = 10^(-pH) = 10^(-3.3425) = 4.5446*10-4 ≈ 4.54*10-4 M (ans).

4) We have

2 H2O (l) ------> H3O+ (aq) + OH- (aq)

Given, [OH-] = 2.5*10-6 M, we know that [H3O+][OH-] = Kw where KW = 1.0*10-14. Plug in values to obtain

[H3O+](2.5*10-6) = 1.0*10-14

====> [H3O+] = 1.0*10-14/(2.5*10-6) = 4.0*10-9 M (ans).


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