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For a solution that contains 0.10 M HCN and 0.10 M NaCN, whi

  • #1

Homework Statement



>For a solution that contains 0.10 M HCN and 0.10 M NaCN, which statement is false?
A) This is an example of the common ion effect.
B) The [H+] is larger than it would be if only the HCN was in solution.
C) the [H+] is equal to $K_a$
D) Addition of more NaCN will shift equilibrium to the left
E) Addition of NaOH will increase [CN^-] and decrease [HCN]

Homework Equations




The Attempt at a Solution


I was actually stumped when I first saw the question. I tried wirintg the equation HCN + NaOH --> NaCN + H2O, and made an ice table so to get (0.1 + x)x/(0.1-x), but I wasn't sure where to go after that.
 

Answers and Replies

  • #2
Borek
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No need for any calculations here, it is about understanding what is going on.

Can you name the acid and the conjugate base? What is ratio of their concentrations?
 
  • #3
@Borek — The acid is HCN and the conj base NaCN, and the concentrations have ratios of 1:1
 
  • #4
epenguin
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This is an important special case. You have mentioned various ions. Can you write an equation for the reaction Involving the ions that you have mentioned? Then an expression for the Ka that you have mentioned.
 
  • #5
@epenguin
I think HCN + NaOH --> H2O + NaCN could work.

then Ka = [H2O][0.10] / [NaOH][0.10] = [H2O]/[NaOH].


so,
for (A), I'm not sure what the common ion effect is so I can't say.
(B) I'm not sure what this is talking about. But I think [H+] would be larger when only HCN is in the solution (bc then there won't be any bases like NaOH), if I understand it right, so this looks like it could be false.
(C) I can't say bc of the [NaOH] in the denominator.
(D) yes, by Le Chatelier's principle, so rule this out.
from the equation, for (E), the addition will shift equilibrium to right so [NaCN] increases and the statement is true, so eliminate this.
 
  • #6
epenguin
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There are no ions in your #5. Yet they appear in #1
 
  • #7
what do you mean by #5 and #1?
 
  • #8
epenguin
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Look at the blue strip above this text. This is #8
 
  • #9
Borek
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