A certain indicator HIn has a \(\mathrm{p} K_{2}\) of \(3.00\) and a color change becomes visible when \(7.00 \%\) of the indicator has been converted to \(\mathrm{In}^{-}\). At what \(\mathrm{pH}\) is this color change visible?

In any aqueous solution, the relationship between the acidity (pH) and basicity (pOH) can be expressed as: \(pH + pOH = 14\) For our problem, we will focus on the pH, which is representative of the acidity of the solution.

We are given that 7.00% of the indicator has been converted to In⁻. If we denote the full amount of HIn as 1, then the amounts of HIn and In⁻ will be as follows: Amount of HIn = 0.93 (since 100% - 7% = 93%) Amount of In⁻ = 0.07 (from information given)

We are given that the pK₂ of HIn is 3.00. The pK₂ represents the acidity constant for the second dissociation step. It is related to the equilibrium constant (K₂) by the equation: \[ pK_{2} = -\log K_{2} \] We can now calculate K₂: \[ K_{2} = 10^{-pK_{2}} = 10^{-3.00} = 1.00 \times 10^{-3} \]

The equilibrium expression for the dissociation of HIn to In⁻ is given by: \[ K_{2} = \frac{[\mathrm{In}^{-}] [\mathrm{H}^{+}]}{[\mathrm{HIn}]} \] Substituting K₂, the amounts of In⁻ and HIn, and denoting the concentration of H⁺ ions as [H⁺], we have: \[ 1.00 \times 10^{-3} = \frac{(0.07) ([\mathrm{H}^{+}])}{(0.93)} \] Now we can solve for the concentration of H⁺ ions: \[ [\mathrm{H}^{+}] = \frac{1.00 \times 10^{-3} \times 0.93}{0.07} \] \[ [\mathrm{H}^{+}] = 1.33 \times 10^{-2} \] Finally, we can convert the concentration of H⁺ ions to pH using the equation: \[ pH = -\log [\mathrm{H}^{+}] \] \[ pH = -\log (1.33 \times 10^{-2}) \] Therefore, the pH at which the color change becomes visible is approximately 1.88.