Finding pH of a solution

[americanpierg]

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1.

From destroyer, find the pH of a .040 M NaCN solution, given that Ka of HCN is 5.0 x 10^-10

Do you necessarily need to find the Kb like they did, or can you just use the Ka and solve the equilibrium equation with that? I understand using Kb is the "politically correct" way to do the problem since its a base dissociation, but doing it both way gets the same answer for this problem.

Using Kb, you get pOH as 3, Using Ka, you get pH as 11. Was this just a concidence that the answer was the same or does it not matter whether you use Ka or Kb becuase answer will be the same?


ALSO

2

If you're given a diprotic acid, what would the equilibrium eqation be? Say the word problem says Phosphoric Acid dissociates in water, and Ka is Y.

Is the equation [H+][H2PO42-]/[H3PO4] or [3H+][PO43-]/[H3PO4] ???? Will they give you all the Ka's.

Any help will be greatly appreciated. Thanks

 

[wantVCUdental]

For the first part, i think Ka and Kb are both appropriate for calculation. you get the same answer, because Ka and Kb are inter-related (if you got different answer then you probably did it wrong eh he)

for the second part about diprotic acids, i don't think they will ask you about it. However, in the case that they come up, you'll be using both constants and usually, the first proton has to deprotonate on all the existing species in the solution before the second proton is removed
so you would use the Keq for the first proton and if there's still more base in the solution, you start using the Keq for second proton.

 

[chromosome21]

1. answer has to be the same, it doesn't matter whether you use Ka or Kb values, just make sure with Ka you use Ka values not Kb values cause Ka*Kb = 1x10^-14. So if the Ka is given, first find out pH and b/c pH+pOH = 14, you can then find out what pOH is. The answer will be the same if you initially convert Ka value into Kb value by using the equation I mentioned before, because then you'll directly get the concentration of the base (here, CN-), and you can use pOH = -log[base]. In short, the answer will be the same. Remember, Ka*Kb = 1x10^-14; and pH + pOH = 14.

2. I'd say the equation would be [H+][H2PO42-]/[H3PO4] because all the Ka values are for the first dissociation of hydrogen. I'm not 100% positive though, I still have to study g.chem 🙂.

1.

From destroyer, find the pH of a .040 M NaCN solution, given that Ka of HCN is 5.0 x 10^-10

Do you necessarily need to find the Kb like they did, or can you just use the Ka and solve the equilibrium equation with that?

Using Kb, you get pOH as 3, Using Ka, you get pH as 11. Was this just a concidence that the answer was the same or does it not matter whether you use Ka or Kb becuase answer will be the same?


ALSO

2

If you're given a diprotic acid, what would the equilibrium eqation be? Say the word problem says Phosphoric Acid dissociates in water, and Ka is Y.

Is the equation [H+][H2PO42-]/[H3PO4] or [H+][PO43-]/[H3PO4] ????

Any help will be greatly appreciated. Thanks

 

[americanpierg]

1. answer has to be the same, it doesn't matter whether you use Ka or Kb values, just make sure with Ka you use Ka values not Kb values cause Ka*Kb = 1x10^-14. So if the Ka is given, first find out pH and b/c pH+pOH = 14, you can then find out what pOH is. The answer will be the same if you initially convert Ka value into Kb value by using the equation I mentioned before, because then you'll directly get the concentration of the base (here, CN-), and you can use pOH = -log[base]. In short, the answer will be the same. Remember, Ka*Kb = 1x10^-14; and pH + pOH = 14.

2. I'd say the equation would be [H+][H2PO42-]/[H3PO4] because all the Ka values are for the first dissociation of hydrogen. I'm not 100% positive though, I still have to study g.chem 🙂.

Do you mean concentration of OH- when you said that? Sorry, I've got every concept of GChem down (25 on first achiever exam) but I never paid attention on Acids/Bases and these concentration equations as well as buffer are killing me.

Equation is CN- + H2O ---> HCN + OH- btw,

so kb = [HCN][OH-]/[CN-]

thats why they found kb first, but instead of finding kb, i just plugged ka into where kb was supposed to be, but same answer. does it still not make a difference?

 

[chromosome21]

oh I thought it was HCN ---> H+ + CN-, so in this case CN- is the base.

Do you mean concentration of OH- when you said that? Sorry, I've got every concept of GChem down (25 on first achiever exam) but I never paid attention on Acids/Bases and these concentration equations as well as buffer are killing me.

Equation is CN- + H2O ---> HCN + OH- btw,

so kb = [HCN][OH-]/[CN-]

thats why they found kb first, but instead of finding kb, i just plugged ka into where kb was supposed to be, but same answer. does it still not make a difference?

 

[chromosome21]

Plus, you add NaCN, which is a salt, so it'll dissociate completely and CN- concentration in the solution would be x+.4 (or was it .04?), it's a common ion effect, i think.

HCN = H+ + CN-
-x +x +x + [common ion CN- that you get from the salt]
+x+.4
b/c HCN is a very weak acid, it will not dissociate completely, and at the equilibrium it'll be nearly the same as the initial concentration.

now, Ka = [H+] * [CN-]

5.0x10^-10 = x* [x+0.4]
b/c x<<<.4, so that we can ignore the x for CN-
so, 5.0x10^-10 = x* [0.4]
therefore, x = 1. * 10^-11
remember, we ignored x from the CN- concentration b/c adding this answer with 0.4 would nearly be 0.4.
Now, the x is the concentration of H+ ions in the solution here.
so, pH = -log [H+]
pH = - log [10^-11]
pH = 11
pOH = 14-pH
pOH = 3

Do you mean concentration of OH- when you said that? Sorry, I've got every concept of GChem down (25 on first achiever exam) but I never paid attention on Acids/Bases and these concentration equations as well as buffer are killing me.

Equation is CN- + H2O ---> HCN + OH- btw,

so kb = [HCN][OH-]/[CN-]

thats why they found kb first, but instead of finding kb, i just plugged ka into where kb was supposed to be, but same answer. does it still not make a difference?

 

[chromosome21]

oh I misread you. you're saying you plugged in Ka value instead of Kb value and the answer was the same? are you sure it's a Ka value? it might be a typo, i don't have destroyer, so can't help you with that, but if there is Kb, it has to be a kb value not ka value. if ka value is given, you have to convert it into kb value. Please correct me if i am wrong.

Do you mean concentration of OH- when you said that? Sorry, I've got every concept of GChem down (25 on first achiever exam) but I never paid attention on Acids/Bases and these concentration equations as well as buffer are killing me.

Equation is CN- + H2O ---> HCN + OH- btw,

so kb = [HCN][OH-]/[CN-]

thats why they found kb first, but instead of finding kb, i just plugged ka into where kb was supposed to be, but same answer. does it still not make a difference?

 

[Nasem]

Plus, you add NaCN, which is a salt, so it'll dissociate completely and CN- concentration in the solution would be x+.4 (or was it .04?), it's a common ion effect, i think.

HCN = H+ + CN-
-x +x +x + [common ion CN- that you get from the salt]
+x+.4
b/c HCN is a very weak acid, it will not dissociate completely, and at the equilibrium it'll be nearly the same as the initial concentration.

now, Ka = [H+] * [CN-]

5.0x10^-10 = x* [x+0.4]
b/c x<<<.4, so that we can ignore the x for CN-
so, 5.0x10^-10 = x* [0.4]
therefore, x = 1. * 10^-11
remember, we ignored x from the CN- concentration b/c adding this answer with 0.4 would nearly be 0.4.
Now, the x is the concentration of H+ ions in the solution here.
so, pH = -log [H+]
pH = - log [10^-11]
pH = 11
pOH = 14-pH
pOH = 3

I know this problem from destroyer... its actually [0.04] initial concentration of NaCN.... so if you plugged in 0.04 instead of the 0.4, you would not get Ph of 11... instead u'd get pH of 8

here is what I mean...

Initially CN- + water <--> HCN + OH-

Okay, so you wanna use the Ka (which is given in problem)

HCN + water <--> CN- + Acid , where Ka = 5 * 10^-10

so, since the molar solubility of HCN is X, you have one X from CN- and X from acid portion..... to make things easier for claculations, the X from CN- can be neglected and replaced by 0.04, so

Ka = 5 * 10^-10 = (0.04) (X)
simplify this...
1 * 10^-8 = X
so...

pH = -log(10^-8) = 8


I am a bit confused as to how you guys are getting pH of 11 if you use either the Ka or Kb, clearly the answer is 11, but I can only get it by using Kb, NOT ka

 

[wantVCUdental]

Your logic is flawed here, if you use Ka, then the variable x must represent acid if you want to use pH=-log[x]
however, in the equation you've written, x represents moles of HCN and OH-.
Now you can conveniently calculate for pH just bying knowing the concentration of OH-, i don't know why you guys are adamant about using Ka, just use Kb, with OH- there, pH is easily solved.

good luck

I know this problem from destroyer... its actually [0.04] initial concentration of NaCN.... so if you plugged in 0.04 instead of the 0.4, you would not get Ph of 11... instead u'd get pH of 8

here is what I mean...

Initially CN- + water <--> HCN + OH-

Okay, so you wanna use the Ka (which is given in problem)

HCN + water <--> CN- + Acid , where Ka = 5 * 10^-10

so, since the molar solubility of HCN is X, you have one X from CN- and X from acid portion..... to make things easier for claculations, the X from CN- can be neglected and replaced by 0.04, so

Ka = 5 * 10^-10 = (0.04) (X)
simplify this...
1 * 10^-8 = X
so...

pH = -log(10^-8) = 8


I am a bit confused as to how you guys are getting pH of 11 if you use either the Ka or Kb, clearly the answer is 11, but I can only get it by using Kb, NOT ka

 

[Nasem]

Your logic is flawed here, if you use Ka, then the variable x must represent acid if you want to use pH=-log[x]
however, in the equation you've written, x represents moles of HCN and OH-.
Now you can conveniently calculate for pH just bying knowing the concentration of OH-, i don't know why you guys are adamant about using Ka, just use Kb, with OH- there, pH is easily solved.

good luck

wantVCUdental,

Im sorry, I don't follow you,

can you please show how you would do it using the Ka (I know, going with Kb is easier, but its fun knowing how to do it using Ka)

 

[americanpierg]

wantVCUdental,

Im sorry, I don't follow you,

can you please show how you would do it using the Ka (I know, going with Kb is easier, but its fun knowing how to do it using Ka)

Lol, you can't do it using Ka guys... well I guess you can but it would be too diificult to even try. When I said Ka I just meant using the same equation, but switching Ka for Kb. I got 1x10^-11 but forgot to square root each side. anways

Just think about it. If you switch the equations backwards, the equilibrium equation

HCN--> CN + H

If you use 0.04M as the concentration of CN, you will have to work backwards to find the HCN since you were never given a concentration of HCN. Working backwards... it would just be

CN+H2O---> HCN... equation we started with.


Solubility doesn't play a role in this...