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

In summary, the statement that is false is that the [H+] is larger than it would be if only the HCN was in solution.
  • #1
JessicaHelena
188
3

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.
 
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  • #2
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
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
There are no ions in your #5. Yet they appear in #1
 
  • #7
what do you mean by #5 and #1?
 
  • #8
Look at the blue strip above this text. This is #8
 
  • #9
JessicaHelena said:
what do you mean by #5 and #1?

Post number in the thread.
 

1. What is the pH of the solution?

The pH of the solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of HCN, [A-] is the concentration of the conjugate base (CN-) and [HA] is the concentration of the acid (HCN). In this case, the pH would be approximately 9.24, assuming a pKa of 9.24 for HCN.

2. Is the solution basic or acidic?

The solution is basic, with a pH above 7. This is because the concentration of the conjugate base (CN-) is equal to the concentration of the acid (HCN), resulting in a solution with a pH greater than the pKa of HCN.

3. What is the concentration of CN- in the solution?

The concentration of CN- can be calculated using the dissociation constant (Ka) of HCN and the initial concentrations of HCN and NaCN. Using the formula Ka = [H+][CN-]/[HCN], the concentration of CN- can be determined to be 0.05 M.

4. How does the concentration of NaCN affect the pH of the solution?

The concentration of NaCN does not directly affect the pH of the solution. However, it does affect the concentration of CN-, which in turn affects the pH of the solution through the Henderson-Hasselbalch equation.

5. What is the significance of having equal concentrations of HCN and NaCN in the solution?

The equal concentrations of HCN and NaCN result in a buffer solution, which is resistant to changes in pH. This is because the weak acid (HCN) and its conjugate base (CN-) can neutralize any added acid or base, maintaining the pH of the solution. This makes it useful in many industrial and biological processes where a stable pH is necessary.

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