A \(0.050-M\) solution of the salt \(\mathrm{NaB}\) has a pH of \(9.00\). Calculate the \(\mathrm{pH}\) of a \(0.010-M\) solution of \(\mathrm{HB}\).

Both NaB and HB are part of the same acid-base system, where they are the conjugate base and acid, respectively. The equilibrium involved in this system is represented as: \[ \mathrm{HB} \rightleftharpoons \mathrm{H^+} + \mathrm{B^-} \] It is important to remember that NaB dissociate completely into Na+ and B- ions in the solution.

Given the concentration of NaB and the pH, we can calculate the concentration of the conjugate base (\(\mathrm{B^-}\)) and the protonated acid (\(\mathrm{HB}\)) species: \[ \mathrm{[NaB]} = 0.050\: M \] Since NaB dissociates completely, the concentration of B- ions in the solution is equal to the concentration of NaB: \[ \mathrm{[B^-]} = 0.050\: M \] As the pH is given as 9.00, we can find the concentration of H+ ions: \[ \mathrm{[H^+]} = 10^{-\mathrm{pH}} = 10^{-9.00} = 1.0 \times 10^{-9} \: M \] From the equilibrium expression, we can now find the concentration of HB in the base solution: \[ \mathrm{[HB]} = \frac{\mathrm{[H^+]} \times \mathrm{[B^-]}}{\mathrm{[B^-]}} = \frac{(1.0 \times 10^{-9}\: M) \times (0.050\: M)}{0.050\: M} = 1.0 \times 10^{-9}\: M \]

Now that we have the concentrations of HB, H+, and B-, we can calculate the acid-dissociation constant (Ka) for boric acid: \[ K_{a} = \frac{\mathrm{[H^+]} \times \mathrm{[B^-]}}{\mathrm{[HB]}} = \frac{(1.0 \times 10^{-9}\: M) \times (0.050\: M)}{(1.0 \times 10^{-9}\: M)} = 5 \times 10^{-8} \]

With the Ka value known, we can now solve for the pH of the 0.010 M HB solution. We start by setting up the equilibrium equation for the given concentration of boric acid: \[ K_{a} = \frac{\mathrm{[H^+]} \times \mathrm{[B^-]}_{0.010-M \: solution}}{\mathrm{[HB]}_{0.010-M \: solution}} \] We can assume that, at equilibrium, the concentrations of all species are: \[\mathrm{[HB]}_{0.010-M \: solution} = (0.010 - x)\: M\] \[\mathrm{[H^+]}_{0.010-M \: solution} = x\: M \] \[\mathrm{[B^-]}_{0.010-M \: solution} = x\: M \] Substitute these into the Ka expression calculated earlier: \[ 5 \times 10^{-8} = \frac{x^{2}}{(0.010 - x)} \] Since Ka value is very small, we can make the approximation that x is also very small and hence can be neglected in comparison with 0.010. So, the equation becomes: \[ 5 \times 10^{-8} = \frac{x^{2}}{(0.010)} \] Solve for x: \[ x = \sqrt{5 \times 10^{-8} \times 0.010} = 7.071 \times 10^{-5} M \] Since x is the concentration of H+ ions, we can calculate the pH of the solution as: \[ \mathrm{pH} = -\log{(7.071 \times 10^{-5})} \approx 4.150 \] The pH of the 0.010 M HB solution is approximately 4.150.