Calculate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%
Calculate the Ka of nitrous acid if a 0.200 mol/L solution at equilibrium at SATP has a percent ionization of 5.8%
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chemistry - DBob222, Saturday, April 5, 2008 at 4:50pm
HNO2 ==> H^+ + NO2^-
Initial:
(HNO2) = 0.2
(H^+) = 0
(NO2^-) = 0
Ka = (H^+)(NO2^-)/(HNO2)
If 5.8% ionized, then after ionization:
(H^+) = 0.2 x 0.058 = ??
(NO2^-) = 0.2 x 0.058 = ??
(HNO2) = 0.2 x (1-0.058) = ?? (Note: If soln is 5.8% ionized than unionized is 100 - 5.8 = 94.2% OR 1.00 - 0.058 = 0.942).
Plug these values into Ka expression above and solve for Ka.
Here is my work:
Ka = (H^+)(NO2^-)/(HNO2)
= (0.0116)(0.0116) / (0.1884)
= 1.34 X 10^-4 / (0.1884)
= 7.14 X 10^-4
My answer is still wrong:( Is there a step I missed?
HNO2 ==> H^+ + NO2^-
Initial:
(HNO2) = 0.2
(H^+) = 0
(NO2^-) = 0
Ka = (H^+)(NO2^-)/(HNO2)
If 5.8% ionized, then after ionization:
(H^+) = 0.2 x 0.058 = ??
(NO2^-) = 0.2 x 0.058 = ??
(HNO2) = 0.2 x (1-0.058) = ?? (Note: If soln is 5.8% ionized than unionized is 100 - 5.8 = 94.2% OR 1.00 - 0.058 = 0.942).
Plug these values into Ka expression above and solve for Ka.
I'm doing this exact same question lol, I got 7.1x10^-4
Well, let me dip my toes into the world of chemistry here. It sounds like you want to calculate the Ka of nitrous acid based on its percent ionization. Nitrous acid, HNO2, can dissociate into H+ ions and NO2- ions.
So, if the percent ionization is 5.8%, it means that 5.8% of the initial nitrous acid has ionized. That also means that 94.2% of it remains in the solution.
Now, let's call the initial concentration of nitrous acid [HA]. After ionization, the concentration of H+ ions and NO2- ions will both be equal to 0.058[HA] (since they were formed in equal amounts). And the concentration of remaining nitrous acid will be 0.942[HA].
Now, let's write the equation for the dissociation of nitrous acid:
HNO2 ⇌ H+ + NO2-
The equilibrium expression is:
Ka = [H+][NO2-]/[HA]
Substituting the concentrations we found earlier, we get:
Ka = (0.058[HA])(0.058[HA])/[0.942[HA]]
Simplifying, we have:
Ka = 0.058^2 / 0.942
And that, my friend, is how you calculate the Ka of nitrous acid with a 0.200 mol/L solution at equilibrium with a percent ionization of 5.8%.
To calculate the acid dissociation constant (Ka) of nitrous acid (HNO2) given the percent ionization, you first need to determine the concentration of the dissociated acid and undissociated acid at equilibrium.
Let's define the following variables:
- [HNO2]: Concentration of undissociated nitrous acid
- [H+]: Concentration of hydrogen ions (from dissociated nitrous acid)
- [NO2-]: Concentration of nitrite ions (from dissociated nitrous acid)
- Initial concentration of nitrous acid (HNO2): 0.200 mol/L
Given that the percent ionization is 5.8%, this means that 5.8% of the initial nitrous acid has dissociated. Therefore, the concentration of hydrogen ions and nitrite ions can also be calculated.
Percent ionization = (concentration of dissociated acid / initial concentration of acid) x 100
5.8% = ([H+] + [NO2-]) / 0.200 mol/L
Since nitrous acid is a monoprotic acid, for every 1 mol of HNO2 that dissociates, you will get 1 mol of H+ and 1 mol of NO2-. Therefore, [H+] and [NO2-] are equal.
Let x be the concentration of H+ and NO2-. Thus, x = [H+] = [NO2-].
Therefore, 5.8% = (2x) / 0.200 mol/L (using the fact that 2x represents the total concentration of the dissociated acid)
Now, let's solve for x:
0.058 = (2x) / 0.200
Cross-multiplying, we get:
0.058 * 0.200 = 2x
0.0116 = 2x
x = 0.0116 / 2
x = 0.0058 mol/L
Therefore, the concentration of hydrogen ions and nitrite ions at equilibrium is 0.0058 mol/L.
Now that we have the values of [H+] and [NO2-], we can calculate the value of Ka.
Ka = ([H+] * [NO2-]) / [HNO2]
Substituting the known values:
Ka = (0.0058 * 0.0058) / 0.200
Ka = 0.00003364 / 0.200
Ka = 0.0001682
So, the Ka of nitrous acid (HNO2) is approximately 0.0001682.