a 0.250 mol sample of HX is dissolved in enough water to form 655mL of solution. If the pH of the solution is 3.44, what is the Ka of HX.
(HX) = moles/L = 0.250/0.655 = approx 0.4 but you need a better answer than that.
.............HX ==> H^+ + X^-
initial.....0.4M.....0......0
change.......-x......x......x
equi.....0.4-x.......x.......x
Convert pH 3.44 to H^+ from
pH = 3.44 = -log(H^+)
That will be x, substitute into the Ka expression and solve for Ka.
Ah, the elusive Ka of HX. It's like trying to find a clown in a circus tent, isn't it? Well, fear not, for Clown Bot is here to help!
To find the Ka of HX, we first need to find the concentration of HX in the solution. We can use the equation C = n/V, where C is the concentration, n is the number of moles, and V is the volume in liters. So, let's get to the calculations!
We have a 0.250 mol sample of HX dissolved in 655 mL of water. To convert the volume to liters, we divide by 1000: 655 mL = 0.655 L.
Now we can find the concentration: C = 0.250 mol / 0.655 L ≈ 0.3818 mol/L.
Since the solution has a pH of 3.44, we can use the formula pH = -log[H+]. We know that [H+] = [HX], as HX is the only acid in this solution. So, we can rewrite the equation as pKa = -log(Ka) = 3.44.
Now, let's find Ka by taking the antilog of both sides: Ka = 10^(-pKa).
Substituting the given value, we get Ka = 10^(-3.44).
Calculating this with the magic of Clown Math, we find that the Ka of HX is approximately 3.52542 x 10^(-4). Voila!
Remember, Clown Bot is not responsible for any laughter or clown sightings that may occur as a result of these calculations. Proceed with caution!
To find the Ka of HX, we need to use the concentration of HX and the pH of the solution. The equation for Ka is:
Ka = [H+] * [X-] / [HX]
First, we need to find the concentration of H+ in the solution. The pH is given as 3.44, which means the concentration of H+ is 10^(-pH). Therefore, [H+] = 10^(-3.44).
Next, we need to find the concentration of HX. We are given that there is a 0.250 mol sample of HX dissolved in 655 mL of solution. To convert this to concentration, we divide the number of moles by the volume in liters:
[HX] = 0.250 mol / 0.655 L
Now that we have [H+] and [HX], we can substitute these values into the Ka equation:
Ka = [H+] * [X-] / [HX]
However, in this case, since HX is a weak acid, we assume that it fully dissociates into H+ and X-. Therefore, [HX] can be considered equal to zero.
Hence, the equation becomes:
Ka = [H+] * [X-] / 0
Since [HX] = 0, the denominator is 0, meaning the Ka value is undefined.
[HX] = .250moles/(655/1000L)
=.38M
HX<=====>H+ + x-
I .38 0 0
C -X +x +x
E .38-x x x
pH = 3.44 to find H+
-log[H+] = pH = H= 3.63x 10^-4
Ka = [H+][x-]/[HX]
= [3.63x10^-4]^2/3.8x10^-1
= 3.5x10^-9