Systematic treatment of equilibrium

[reminiscent]

Homework Statement

Determine [Zn2+], [CN–], and [HCN] in a saturated solution of Zn(CN)₂ with a fixed pH of 1.060. The Ksp for Zn(CN)₂ is 3.0 × 10–16. The K_a for HCN is 6.2 × 10–10.

Homework Equations

The Attempt at a Solution

I don't have a phone or camera right now to take a picture of my work, but here's my steps summarized:
1. I found that the [H+]= 0.0871 M.
2. I used that and the Kw equation to find [OH-] = 1.15 x 10^-13 M.
3. I used the mass balance equation to find a new equation for [HCN].
4. I plugged that into the Kb equation, along with the [OH-] in order to find [CN-].
5. I found [CN-] = 4.31 x 10^-24, then plugged that into the Ksp equation to find [Zn2+] = 1.61 x 10^31 and [HCN] = 3.22 x 10^31. The system said my answers were incorrect.
I tried another way using the ICE method and not using systematic treatment of equilibrium, and I found [Zn2+] = 0.2455 M, [CN-] = 0.491 M, and [HCN] = 68.98 x 10^6 M, but the system still said it was wrong.

Please help. ):

 

[Borek]

Why do you use Kb and [OH^-] instead of just converting Kb to Ka?

3. I used the mass balance equation to find a new equation for [HCN].

How?

The system said my answers were correct.

Or incorrect?

In general once you convert Kb to Ka you will end with three equations in three unknowns.

 

[reminiscent]

Why do you use Kb and [OH^-] instead of just converting Kb to Ka?
How?
Or incorrect?

In general once you convert Kb to Ka you will end with three equations in three unknowns.

Sorry, I will edit my post for certain mistakes...
They gave us the Ka for HCN is 6.2 × 10–10.
They converted it to Kb. I don't understand what you mean, though.
For the mass balance equation, I just did: [HCN] = 2[Zn2+] - [CN-].
I'm so stuck, I don't understand how to correctly find it.

 

[Borek]

If you are given Ka and pH, there is no need to go through the reaction between CN^- and water,

HCN ↔ H⁺ +CN^-

is perfectly enough to describe the HCN/CN^- equilibrium.

Write Ka - this will be your first equation (with two unknowns, as pH is fixed).

Second equation is the mass balance (but it can be simplified - see below).

Third equation is Ksp.

That gives three equations in three unknowns, the rest is just a math. HCN is a very weak acid and it is protonated almost quantitatively at so low pH (so [HCN] >> [CN^-] and the latter can be ignored in the mass balance).

 

[reminiscent]

If you are given Ka and pH, there is no need to go through the reaction between CN^- and water,

HCN ↔ H⁺ +CN^-

is perfectly enough to describe the HCN/CN^- equilibrium.

Write Ka - this will be your first equation (with two unknowns, as pH is fixed).

Second equation is the mass balance (but it can be simplified - see below).

Third equation is Ksp.

That gives three equations in three unknowns, the rest is just a math. HCN is a very weak acid and it is protonated almost quantitatively at so low pH (so [HCN] >> [CN^-] and the latter can be ignored in the mass balance).

Okay, I understand - so the equations are:
[HCN] = 2[Zn2+]
Ka = [OH-][HCN]/[CN-] = 6.2 x 10^-10
Ksp = [Zn2+][CN-]^2 = 3.0 x 10^-16

Which one should I solve for first? I think the part of solving is what confuses me.

 

[reminiscent]

If you are given Ka and pH, there is no need to go through the reaction between CN^- and water,

HCN ↔ H⁺ +CN^-

is perfectly enough to describe the HCN/CN^- equilibrium.

Write Ka - this will be your first equation (with two unknowns, as pH is fixed).

Second equation is the mass balance (but it can be simplified - see below).

Third equation is Ksp.

That gives three equations in three unknowns, the rest is just a math. HCN is a very weak acid and it is protonated almost quantitatively at so low pH (so [HCN] >> [CN^-] and the latter can be ignored in the mass balance).

Okay so far, I made the mass balance equation and Ksp equation equal to each other by isolating [Zn2+], and I solved for [HCN]. Can I just plug that back into the Ka equation, then solve for [CN-] this way? Also, [OH-] would be 1.15 x 10^-13 M, correct?

 

[reminiscent]

If you are given Ka and pH, there is no need to go through the reaction between CN^- and water,

HCN ↔ H⁺ +CN^-

is perfectly enough to describe the HCN/CN^- equilibrium.

Write Ka - this will be your first equation (with two unknowns, as pH is fixed).

Second equation is the mass balance (but it can be simplified - see below).

Third equation is Ksp.

That gives three equations in three unknowns, the rest is just a math. HCN is a very weak acid and it is protonated almost quantitatively at so low pH (so [HCN] >> [CN^-] and the latter can be ignored in the mass balance).

I got [CN-] = 4.81 x 10-7, [Zn2+] = 0.00130, and [HCN] = 0.00259 and it is still incorrect...

 

[Borek]

I can confirm it is incorrect - but there is not much I can do not seeing what and how you did.

 

[Charles Link]

I worked this one and it is interesting. I think I got the correct answer. It uses 3 equations and 3 unknowns. Algebraically it is actually kind of simple. In post #5 your $K_a$ is incorrect and $[H+]$ belongs in the equation instead of $[OH-]$. Since you know the pH, you know the numerical value of $[H+]$. The algebra is simple enough that you can use your complete mass balance of equation of post #3. You can eliminate the $[HCN]$ from the 3 equations using the mass balance equation, and eliminate the $[Zn+2]$ as well and solve for $[CN-]$.

 

[reminiscent]

I worked this one and it is interesting. I think I got the correct answer. It uses 3 equations and 3 unknowns. Algebraically it is actually kind of simple. In post #5 your $K_a$ is incorrect and $[H+]$ belongs in the equation instead of $[OH-]$. Since you know the pH, you know the numerical value of $[H+]$. The algebra is simple enough that you can use your complete mass balance of equation of post #3. You can eliminate the $[HCN]$ from the 3 equations using the mass balance equation, and eliminate the $[Zn+2]$ as well and solve for $[CN-]$.

I think I forgot my algebra skills because I got it wrong again...
Okay I tried what you did. I used the mass balance equation of post 3, plugged that into the Ka equation for [HCN], then I simplified more so I get Ka = 0.0871*2*([Zn2+]/[CN-]). I made the Ksp equation equal to [Zn2+], I plugged that into the Ka equation for [Zn2+]. The [CN-] would have a power of 3 to it, and I got [CN-]=0.00438. It is still wrong.

 

[Charles Link]

Your $K_a$ equation is incorrect. Your mass balance is $[CN-]=2[Zn+2]-[HCN]$ and the shortcut version is $[HCN]=2 [Zn+2]$. In any case $[CN-]$ should be in the numerator and doesn't get replaced. You could put either $[HCN]=(2[Zn+2])$ or $[HCN]=(2[Zn+2]-[CN-])$ in the denominator. ($[CN-]$ is very small, but you really don't need any shortcuts for the algebra.) Y(our number .0871 for the $[H+]$ for the pH =1.06 looks correct.) Your $K_a$ equation is also incorrect in post # 5. It is actually a simple equation and should read $K_a= ([H+][CN-])/[HCN]$. That appears where your error originated...

 

[Charles Link]

I added a couple of things to post #11. Please be sure to read the latest edited version.

 

[reminiscent]

I added a couple of things to post #11. Please be sure to read the latest edited version.

Thank you so much! I can't believe I didn't catch that. I finally got it correct. Thank you!

 

[Borek]

Thank you so much! I can't believe I didn't catch that. I finally got it correct. Thank you!

In the future please show your work - it is much easier to find a mistake seeing what was done, than guessing out of rather cryptic description.

Plus, when you try to put your work in a readable format you are quite likely to find mistakes by yourself.