Question

The solutions in three test tubes labeled \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) all have the same \(\mathrm{pH}\). The test tubes are known to contain \(1.0 \times\) \(10^{-3} \mathrm{M} \mathrm{HCl}, 6.0 \times 10^{-3} \mathrm{M} \mathrm{HCHO}_{2},\) and \(4 \times 10^{-2} \mathrm{M}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\). Describe a procedure for identifying the solutions.

Short answer

Answer: Test tube A contains HCl, test tube B contains HCHO2, and test tube C contains C6H5NH3+.

Step-by-step solution

Determine the acidic/basic nature of each solution

First, let's identify whether each solution is acidic or basic. An acidic solution has a pH below 7, a basic solution has a pH above 7, and a neutral solution has a pH equal to 7. HCl is a strong acid, HCHO2 is a weak acid, and C6H5NH3^{+} is a cation derived from a weak base. Thus, all these solutions should have a pH less than 7.

Calculate the pH of each solution

Since all the given solutions have the same pH, we should compute their pH values. For the strong acid HCl: pH = -log[H+]= -log(1.0 × 10^(-3)) = 3 For the weak acid HCHO2 and its concentration ([HCHO2]) is 6.0 × 10^(-3) M, using an ICE table and Ka values, we can calculate the pH. The Ka for formic acid (HCHO2) is 1.8 × 10^(-4). Let x be the concentration of H+ ions: Ka = [HCHO2-][H+]/[HCHO2] = x^2/(6.0x10^(-3)-x) Solve for x, and we have [H+] = x, which also has a pH = 3 The cation in C6H5NH3^{+} is derived from a weak base (C6H5NH2), which has a Kb of 4.0 × 10^(-10). Converting the Kb to Ka using the relation Ka = (Kw/Kb) = (10^(-14))/4.0 × 10^(-10), we get a Ka of 2.5 × 10^(-5). Analyzing using the same process as before, we have [H+] = x, which also results in a pH = 3.

Propose a procedure to identify the solutions

Since all three solutions have the same pH of 3, we can't use pH alone to identify them. Therefore, we will use their reactions with appropriate reagents. 1. Add naught solution (Na2CO3) to a small portion of the solution from each test tube. HCl will completely react with it, forming bubbles of CO2, while HCHO2 and C6H5NH3+ will not produce any visible reaction as weak acids don't react with Na2CO3. 2. Next, add small portions each of the remaining solutions (from step 1) to the other solutions which didn't react with Na2CO3. This means that one of the solutions would presumably contain HCHO2, and the other would contain C6H5NH3+. When weak acid and weak base react, they form a salt solution. The C6H5NH3+ will act as a weak acid in this case and will react with the weak base form, C6H5NH2, to form a salt solution (C6H5NH2HCHO2). With these two steps, we can identify which test tube contains which solution. Test tube A contains HCl, test tube B contains HCHO2, and test tube C contains C6H5NH3+.