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.

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

1. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=1.1×10−9 M. 2. Calculate [H3O+]...

1. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=1.1×10−9 M. 2. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=2.4×10−2 M . 3. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=6.2×10−12 M then classify the solutions as acidic or basic.

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq)...

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C. [H^+]= [ClO4^-]= [OH^-]= Is the solution acidic, basic or neutral?

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.

A. Calculate [H3O+] in the following aqueous solution at 25 ?C: [OH?]= 1.6×10?9 M . Express...

A. Calculate [H3O+] in the following aqueous solution at 25 ?C: [OH?]= 1.6×10?9 M . Express your answer using two significant figures. B. Calculate [H3O+] in the following aqueous solution at 25 ?C: [OH?]= 3.0×10?2 M . Express your answer using two significant figures. C. Calculate [H3O+] in the following aqueous solution at 25 ?C: [OH?]= 7.0×10?12 M . Express your answer using two significant figures.

1.Calculate the [H3O+] value of each aqueous solution. aspirin with [OH−]=1.4×10−11M. seawater with [OH−]=1.5×10−6M 2. To...

1.Calculate the [H3O+] value of each aqueous solution. aspirin with [OH−]=1.4×10−11M. seawater with [OH−]=1.5×10−6M 2. To calculate pH and use it as a measure of acidity. What is the pH of an aqueous solution with [H3O+] = 4×10−13M ? Carbonated cola is more acidic than coffee or even orange juice because cola contains phosphoric acid. What is the molar concentration of H3O+ in a cola that has a pH of 3.120?

Calculate the [OH−] of each aqueous solution with the following [H3O+]: 1)stomach acid, 5.0×10−2M. Express your...

Calculate the [OH−] of each aqueous solution with the following [H3O+]: 1)stomach acid, 5.0×10−2M. Express your answer using two significant figures. 2)urine, 3.0×10−6M. Express your answer using two significant figures. 3)orange juice, 1.6×10−4M. Express your answer using two significant figures. 4)bile, 6.0×10−9M.Express your answer using two significant figures.

Calculate the [OH−] of each aqueous solution with the following [H3O+]: stomach acid, 2.0×10−2M Express your...

Calculate the [OH−] of each aqueous solution with the following [H3O+]: stomach acid, 2.0×10−2M Express your answer using two significant figures. urine, 7.0×10−6M Express your answer using two significant figures. orange juice, 2.2×10−4M Express your answer using two significant figures. bile, 6.0×10−9M Express your answer using two significant figures.

calculate the value of [-OH] from the given [H3O+] in each solution and label the solution...

calculate the value of [-OH] from the given [H3O+] in each solution and label the solution as acidic or basic: a) [H3O+]=6.5×10 (negative 10 power)M b) " "=8.6×10 (negative 3rd power)M

Formic acid (HCOOH) is secreted by ants. Calculate [H3O+] for a 4.98E-2 M aqueous solution of...

Formic acid (HCOOH) is secreted by ants. Calculate [H3O+] for a 4.98E-2 M aqueous solution of formic acid (Ka = 1.80E-4).

For each strong base solution, determine [H3O+], [OH−], pH, and pOH. A)8.89×10−3 M LiOH [OH−],[H3O+] =...

For each strong base solution, determine [H3O+], [OH−], pH, and pOH. A)8.89×10−3 M LiOH [OH−],[H3O+] = b)8.89×10−3 M LiOH pH,pOH= C)1.15×10−2 M Ba(OH)2 [OH−],[H3O+]= D)1.15×10−2 M Ba(OH)2 pH,pOH = E)1.8×10−4 M KOH [OH−],[H3O+]= F)1.8×10−4 M KOH pH,pOH = G)4.9×10−4 M Ca(OH)2 [OH−],[H3O+] = H)4.9×10−4 M Ca(OH)2 pH,pOH =