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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.

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 K_a = 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^-] = K_w where K_W = 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).

queen_honey_blossom answered 2 years ago

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