Hey guys! Thanks for any help in advance! I'm back with this problem:
Calculate the percent dissociation for a 1.00 × 10-6 M solution of HCN at 25 °C.
Ka = 4.90 × 10-10.
So my question really is do I have to use water when I calculate the percent dissociation because my teacher said that at a certain point if it is so small you'll have to use water?
Here is my maybe answer:
I did this two ways, 1 with and 1 without water
Way 1 (Without water)
1,00*10^-6M 0M 0M
-x x x
1.00*10^-6-x x x
x^2/1.00*10^-6-x=4.9*10^-10
-x^2-4.9*10^-10+4.9*10^-16=0
x=2.19*10^-8
2.19*10^-8/1.00*10^-6*100=2.19%
Way 2 (With Water)
1.00*10^-6M 1.00*10^-7M 0M
-x x x
1.00*10^-6-x 1.00*10^-7+x x
x(1.00*10^-7+x)/(1.00*10^-6-x)=4.9*10^-10
-x^2-1.0049*10^-7+4.9*10^-16=0
x=(4.66*10^-9/1.00*10^-6)*100=.466%
There is a big difference between those two numbers, so does that mean I did it wrong, or does that mean that you have to factor the value of water in for the H+ ions because the M is so small?
To calculate the percent dissociation of HCN, you don't need to consider the presence of water. The dissociation of HCN in water is described by the following equilibrium reaction:
HCN(aq) ⇌ H+(aq) + CN-(aq)
The equilibrium expression for this reaction is:
Ka = [H+(aq)][CN-(aq)] / [HCN(aq)]
Given that the concentration of HCN is 1.00 × 10^-6 M and the value of Ka is 4.90 × 10^-10, you can use the quadratic equation to solve for the concentration of H+ ions.
Using the expression you have provided:
-x^2 - 4.9 × 10^-10 + 4.9 × 10^-16 = 0
Simplifying it:
x = 2.21 × 10^-8 M
To calculate the percent dissociation, divide the concentration of H+ ions by the initial concentration of HCN and multiply by 100:
Percent dissociation = (2.21 × 10^-8 M / 1.00 × 10^-6 M) × 100 ≈ 0.00221%
Therefore, the correct percent dissociation for this 1.00 × 10^-6 M solution of HCN at 25 °C is approximately 0.00221%.
The discrepancy between the two results you obtained in Way 1 and Way 2 may be due to a calculation error or the incorrect application of the equilibrium expression. Applying the correct equilibrium expression and solving the equation should yield the accurate result.
Great question! In order to calculate the percent dissociation of HCN, you need to consider the reaction of HCN with water. The reason for this is that HCN is a weak acid and it will partially dissociate into H+ ions and CN- ions when it is dissolved in water.
The reaction of HCN with water can be represented as follows:
HCN + H2O ⇌ H3O+ + CN-
In your first approach (without considering water), you assumed that the concentration of H+ ions at equilibrium is equal to x. However, without water, the dissociation of HCN cannot occur. This assumption is not valid because the reaction of HCN with water is necessary for the generation of H+ ions.
In your second approach (with water), you correctly considered the reaction of HCN with water. By using the initial concentration of HCN (1.00 × 10^-6 M) and the concentration of H+ ions formed (1.00 × 10^-7 M), you were able to calculate the concentration of the undissociated HCN (1.00 × 10^-6 - x) and set up the equilibrium expression. Solving this equation would give you the value of x, which represents the concentration of H+ ions, and then you can calculate the percent dissociation.
So, to answer your question, you do need to consider the presence of water when calculating the percent dissociation of HCN. Without water, the dissociation cannot occur, and the assumption of H+ concentration being x would not be valid.
Therefore, your second approach is the correct one to use, and the result of 0.466% is the accurate percentage of dissociation for a 1.00 × 10^-6 M solution of HCN at 25 °C.
I hope this explanation helps clarify the concept! Let me know if you have any further questions.