This one is a gnarly scuzzard!!! Here’s my best guess…
Given a solution mix containing 1M Ba(OH)₂ + excess Zn(OH)₂ … Determine pH of solution at equilibrium.
Ba(OH)₂ < == > Ba⁺² + 2OHˉ ; Ksp = 3 x 10ˉ¹⁵
Zn⁺² + 2OHˉ < == > Zn(OH)₂ ; Kf = 2 x 10⁺¹⁵
………………………………………………………………..
Ba(OH)₂ + Zn⁺² < == > Ba⁺² + Zn(OH)₂ ; K(net) = Ksp ∙ Kf = (3 x 10ˉ¹⁵)(2 x 10⁺¹⁵) = 6
This indicates that all of the OHˉ from Ba(OH)₂ is converted to Zn(OH)₂ and final OHˉ concentration depends upon the natural ionization of Zn(OH)₂ …
Dissociation of Zn(OH)₂ …
Zn(OH)₂ < = = > Zn⁺² + 2OHˉ ; Kd = 1/Kf = 1/(2 x 10¹⁵) = 5 x 10ˉ¹⁶
Kd = [Zn⁺²][OHˉ]² = (x)(2x)² = 4x³
=> x = (Kd/4)¹ʹ³ = (5 x 10ˉ¹⁶/4)¹ʹ³ = 5 x 10ˉ⁶
=> [OHˉ] = 2x = 2(5 x 10ˉ⁶) = 1 x 10ˉ⁵
=> pOH = -log[OHˉ] = -log(10ˉ⁵) = -(-5) = 5
=> pH + pOH = 14 => pH = 14 – pH = 14 – 5 = 9
Disclaimer => I did not look up the behavior of Ba(OH)₂ in the presence of of Zn(OH)₂ … but the formation constant for Zn(OH)₂ indicates that any OHˉ delivered into solution from Ba(OH)₂ ionization would spontaneously combine with the available (excess) Zn⁺² and precipitate leaving the solubility of Zn(OH)₂ the determining factor in OHˉ concentration. However, The above calculation assumes dissociation of pptd Zn(OH)₂ in pure water. The solution at equilibrium would contain Ba⁺² ions, Zn⁺² ions and OHˉ ions from ionization of the ppt’d Zn(OH)₂ possibly making this a common ion problem due to the presence of Zn⁺², but I don’t see that as a factor. I’m open to suggestion.
Determine the pH of a solution prepared from 1.0 L of a 0.10 M solution of Ba(OH)2 and excess Zn(OH)2 (s).
The Ksp of Zn(OH)2 is 3×10^-15 and the Kf of Zn(OH)4^2- is 2×10^15.
9 answers
Doc48--my first look.
I didn't go through the problem but it appears to me you have used Ksp of Ba(OH)2 to be that of Kap for Zn(OH)2 of 3E-15. I don't think the problem gives a Ksp for Ba(OH)2; in any event, however, it is much larger than 3E-15
I didn't go through the problem but it appears to me you have used Ksp of Ba(OH)2 to be that of Kap for Zn(OH)2 of 3E-15. I don't think the problem gives a Ksp for Ba(OH)2; in any event, however, it is much larger than 3E-15
Yes, you're right. it give the OH complex. Duh! Will try again. Thanks.
OK, here’s a rework… Zn⁺² ions in water react to form Zn(OH)₂, a gelatinous white solid. Further addition of a strong base (in this problem, Ba(OH)₂) reacts with the Zn(OH)₂ to form Zn(OH)₄²ˉcomplex that is soluble in the aqueous base solution. For addition of 2M Ba(OH)₂ to a solution of Zn(OH)₂ …
2M Ba(OH)₂ + Excess Zn(OH)₂ => 2M Zn(OH)₄²ˉ + 2M Ba⁺²
…….Zn(OH)₄²ˉ <==> Zn⁺² + 4OHˉ; Kd = 1/Kf = 1/(2 x 10¹⁵) =5 x 10ˉ¹⁶
Eq ….2M………….….x…......4x
Kd = [Zn⁺²][OHˉ]⁴/[Zn(OH)₄²ˉ]
5 x 10ˉ¹⁶ = (x)(4x)⁴/(2) = 256x⁵/2 = 128x⁵ => x⁵ = 5 x 10ˉ¹⁶/128 = 3.9 x 10ˉ¹⁸
X = 5th root of ( 3.9 x 10ˉ¹⁸) = 3.3 x 10ˉ⁴
[OHˉ] = 2x = 2(3.3 x 10ˉ⁴) = 6.6 x 10ˉ⁴
pOH = - log[OHˉ] = log(6.6 x 10ˉ⁴) = -(-3.2) = 3.2
pH = 14 – pOH = 14 – 3.2 = 10.8
2M Ba(OH)₂ + Excess Zn(OH)₂ => 2M Zn(OH)₄²ˉ + 2M Ba⁺²
…….Zn(OH)₄²ˉ <==> Zn⁺² + 4OHˉ; Kd = 1/Kf = 1/(2 x 10¹⁵) =5 x 10ˉ¹⁶
Eq ….2M………….….x…......4x
Kd = [Zn⁺²][OHˉ]⁴/[Zn(OH)₄²ˉ]
5 x 10ˉ¹⁶ = (x)(4x)⁴/(2) = 256x⁵/2 = 128x⁵ => x⁵ = 5 x 10ˉ¹⁶/128 = 3.9 x 10ˉ¹⁸
X = 5th root of ( 3.9 x 10ˉ¹⁸) = 3.3 x 10ˉ⁴
[OHˉ] = 2x = 2(3.3 x 10ˉ⁴) = 6.6 x 10ˉ⁴
pOH = - log[OHˉ] = log(6.6 x 10ˉ⁴) = -(-3.2) = 3.2
pH = 14 – pOH = 14 – 3.2 = 10.8
correction ... [OH] = 4x, not 2x ...
[OHˉ] = 4x = 4(3.3 x 10ˉ⁴) = 1.3 x 10ˉ³
pOH = - log[OHˉ] = log(1.3 x 10ˉ³) = -(-2.9) = 2.9
pH = 14 – pOH = 14 – 2.9 = 11.1
[OHˉ] = 4x = 4(3.3 x 10ˉ⁴) = 1.3 x 10ˉ³
pOH = - log[OHˉ] = log(1.3 x 10ˉ³) = -(-2.9) = 2.9
pH = 14 – pOH = 14 – 2.9 = 11.1
I know its been a year, but why is the Zn(OH)4^2- concentration 2M? Shoudn't it be 0.10 M, like the initial Ba(OH)2 concentration?
yea idk why but why I try it with the 0.1 I get a totally different pH. Do you know why they have it as 2M?
Zn(OH)₂ (s) ⇌ Zn²⁺ (aq) + 2OH⁻ (aq) Kₛₚ = 3×10⁻¹⁵
Zn²⁺ (aq) + 4OH⁻ (aq) ⇌ Zn(OH)₄²⁻ (aq) Kf = 2×10¹⁵
________________________________
Zn(OH)₂ (s) + 2OH⁻ (aq) ⇌ Zn(OH)₄²⁻ (aq) K = (Kₛₚ)(Kf) = 6
[OH⁻] [Zn(OH)₄²⁻]
I 0.20 0.00
C -2x +x
E 0.20-2x x
K = [Zn(OH)₄²⁻] / [OH⁻]²
6 = x / (0.20 - 2x)
0 = 24x² - 5.8x + 0.24
x = 0.05
[OH⁻] = 0.20 - 2(0.05)
= 0.09
[H₃O⁺] = 1.0×10⁻¹⁴ / 0.09
= 1.06×10⁻¹³
pH = -log[H₃O⁺]
= -log(1.06×10⁻¹³)
= 12.97 (round to 13 for the sake of significant figures)
∴ The pH of the solution is 13.
Zn²⁺ (aq) + 4OH⁻ (aq) ⇌ Zn(OH)₄²⁻ (aq) Kf = 2×10¹⁵
________________________________
Zn(OH)₂ (s) + 2OH⁻ (aq) ⇌ Zn(OH)₄²⁻ (aq) K = (Kₛₚ)(Kf) = 6
[OH⁻] [Zn(OH)₄²⁻]
I 0.20 0.00
C -2x +x
E 0.20-2x x
K = [Zn(OH)₄²⁻] / [OH⁻]²
6 = x / (0.20 - 2x)
0 = 24x² - 5.8x + 0.24
x = 0.05
[OH⁻] = 0.20 - 2(0.05)
= 0.09
[H₃O⁺] = 1.0×10⁻¹⁴ / 0.09
= 1.06×10⁻¹³
pH = -log[H₃O⁺]
= -log(1.06×10⁻¹³)
= 12.97 (round to 13 for the sake of significant figures)
∴ The pH of the solution is 13.
How did you get -5.8x shouldn't it be -4.8x?
1 answer