Captain Kirk, of the Starship Enterprise, has been told by his superiors that only a chemist can be trusted with the combination to the safe containing the dilithium crystals that power the ship. The combination is the pH of Solution A described below, followed by the pH of Solution C. (For example, if the pH of Solution A is 3.47 and the pH of Solution C is 8.15, then the combination to the safe is 3-47-8-15.) The chemist must find the combination using the information below. All solutions are at 25°C. (20 points)
Solution A is 50 mL of 0.1 M solution of the weak monoprotic acid HX.
Solution B is a 0.05 M solution of the salt NaX. It has a pH of 10.02.
Solution C is made by adding 15 mL of 0.25 M KOH to Solution A.
What are you suppose to do with Solution B to help solve for solution A.
My work:
10^-10.02 = 9.55x 10^-11= [H+]
[OH-]= 1.047 x 10^.4
Is 9.55 x 10^-11 the Ka of Solution B.
NaX--Na+ + X
X + H2O-- HX + OH
wat should i do to solve for B to solve for A.
what should I do after this
Responses
Chemistry - DrBob222, Tuesday, May 27, 2008 at 9:02pm
Use solution B to solve for the Ka of the weak acid.
Use Ka from above to solve for the pH of solution A. That gives you the first 3 numbers of the combination.
Use solution C and the Henderson-Hasselbalch equation to solve for the pH of that solution which will be last three numbers of the combination.
Chemistry - Tom, Tuesday, May 27, 2008 at 9:36pm
how do u use Solution B to solve for the Ka?
Solution B is 0.05 M solution of NaX.
The hydrolysis of X^- is
X^- + HOH ==> HX + OH^-
Kb = Kw/Ka = (HX)(OH^-)/(X^-)
You know pH is 10.02. Subtract from 14 to get pOH and solve for (OH^-) remembering that pOH= -log(OH^-). You know (HX) = (OH^-). YOu know Kw and (X^-) so the only unknown is Ka. Solve for Ka.
Uhh, That's... a lot...
To use Solution B to solve for the Ka of the weak acid, you can use the pH of Solution B and the equation for the dissociation of the acid to calculate the concentration of HX (the weak monoprotic acid) in Solution B.
First, recall the equation for the dissociation of the acid:
HX + H2O -> H3O+ + X-
Since the pH of Solution B is given as 10.02, you can calculate the concentration of H3O+ (which is equal to [H+]) using the equation:
[H+] = 10^(-pH)
In this case, [H+] = 10^(-10.02) = 9.55 x 10^(-11) M.
Next, use stoichiometry to determine the concentration of X- in Solution B. From the equation, you can see that for every one molecule of HX that dissociates, one molecule of X- is produced. Therefore, the concentration of X- is equal to the concentration of HX in Solution B.
Finally, use the concentrations of HX and X- to calculate the Ka of the weak acid using the equation:
Ka = [H+] * [X-] / [HX]
Substitute the values you obtained for [H+], [X-], and [HX] into the equation to calculate the Ka of the weak acid in Solution A.
Once you have the Ka of the weak acid, you can use it to solve for the pH of Solution A (which will give you the first three numbers of the combination) using the equation:
pH = -log10(sqrt(Ka * [HX]))
Finally, you can use Solution C and the Henderson-Hasselbalch equation to solve for the pH of that solution, which will give you the last three numbers of the combination.