Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 $\mathrm{M} \mathrm{HONH}_{2}$ by 0.100 $\mathrm{M} \mathrm{HCl} .\left(K_{\mathrm{b}} \text { for } \mathrm{HONH}_{2}=1.1 \times 10^{-8} .\right)$ a. Calculate the \(\mathrm{pH}\) after 0.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. b. Calculate the \(\mathrm{pH}\) after 25.0 \(\mathrm{mL}\) of \(\mathrm{HCl}\) has been added. c. Calculate the \(\mathrm{pH}\) after 70.0 \(\mathrm{mL}\) of HCl has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 300.0 \(\mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the pH = 6.04?

To find the initial pH, we only need to consider the dissociation of HONH2, since no HCl has been added yet. The equilibrium expression is: \(HONH_2 + H_2O \leftrightarrow HONH_3^+ + OH^-\) Using the Kb expression: \(K_b = \frac{[HONH_3^+][OH^-]}{[HONH_2]}\) Since the initial concentration of HONH2 is given as 0.200 M, we will set up an ICE table and solve for x (the change in concentration of the species): \(K_b = \frac{x^2}{0.200 - x}\) Plug in the given Kb value: \(1.1\times10^{-8} = \frac{x^2}{0.200 - x}\) Solve for x (concentration of OH-) and use this to calculate the pH using the formula: \(pOH = -log_{10}([OH^-])\) and \(pH = 14 - pOH\) b. Calculate the pH after 25.0 mL of HCl has been added.

First, we need to find out the new concentration of the base (HONH2) by calculating the moles and concentration of added HCl (0.100 M x 0.025 L). Then, set up a new ICE table with updated initial concentrations and calculate the pH as in part a. c. Calculate the pH after 70.0 mL of HCl has been added.

Repeat the same procedure as in part b, using the new volume of added HCl (0.070 L). d. Calculate the pH at the equivalence point.

At the equivalence point, all HONH2 has been neutralized by HCl. Calculate the volume of HCl needed to neutralize HONH2 completely (100.0 mL x 0.200 M = x L x 0.100 M). Use this volume to calculate the concentration of the HONH3+ ion formed at the equivalence point, and find the pH using the Ka expression for the HONH3+ ion (Ka = Kw / Kb for the conjugate acid-base pair). e. Calculate the pH after 300.0 mL of HCl has been added.

Repeat the procedure as in parts b and c, using the new volume of added HCl (0.300 L). f. At what volume of HCl added does the pH = 6.04?

To find the volume of HCl needed to reach a pH of 6.04, set up an ICE table for a reaction between HONH2 and the HCl added up to that point. The initial concentrations will depend on the moles of each species. Use moles of HCl (moles_HCl = moles_HONH_erased) to find the volume of HCl needed at this point by dividing moles_HCl by the concentration of HCl (0.100 M).