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
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.
how do u use Solution B to solve for the Ka?
Answered above.
To solve for Solution B, you need to determine the concentration of the weak monoprotic acid HX in Solution B. You can use the information given about Solution B's pH to calculate it.
First, calculate the concentration of hydroxide ions ([OH-]) in Solution B using the equation: [OH-] = 10^(-pOH). From the given information, Solution B has a pH of 10.02, so the pOH is 14 - 10.02 = 3.98. Thus, [OH-] = 10^(-3.98).
Since the salt NaX dissociates completely in aqueous solution, the concentration of X- (the conjugate base of the weak acid HX) in Solution B would be the same as the concentration of [OH-]. Therefore, [X-] = [OH-] = 10^(-3.98) M.
To find the concentration of HX in Solution B, you can use the balanced chemical equation X + H2O -> HX + OH-. The stoichiometry of this equation tells us that the concentration of HX is also equal to [OH-].
Therefore, the concentration of HX in Solution B is 10^(-3.98) M.
Now that you have determined the concentration of HX in Solution B, you can use this information to solve for Solution A.
To solve for Solution A, you need to find the pH of Solution A. Solution A is a 50 mL, 0.1 M solution of the weak monoprotic acid HX.
You can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base:
pH = pKa + log([A-]/[HA])
In this case, [A-] represents the concentration of the conjugate base X-, and [HA] represents the concentration of the weak acid HX.
Since Solution B has the same concentration of HX as Solution A, which is 10^(-3.98) M, you can substitute this concentration into the equation and solve for the pH of Solution A.
To calculate the pH of Solution C, which is made by adding 15 mL of 0.25 M KOH to Solution A, you need to consider the reaction between the weak acid HX and strong base KOH.
The reaction between a weak acid and a strong base produces a salt and water. In this case, HX reacts with KOH to form the salt KX and water (H2O). The pH of Solution C would be determined by the concentration of the salt KX.
Since you know the concentration of the salt KX in Solution B is 10^(-3.98) M and the volume of Solution C is the sum of 50 mL (Solution A) + 15 mL (0.25 M KOH), you can calculate the final concentration of the salt KX in Solution C.
With this concentration, you can use the Henderson-Hasselbalch equation again to determine the pH of Solution C, by substituting the concentrations into the equation.
Finally, you can combine the pH of Solution A and the pH of Solution C to form the combination to the safe, as instructed.