Consider the titration of \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HONH \(_{2}\) by \(0.100 \mathrm{M}\) 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 HCI has been added. b. Calculate the \(\mathrm{pH}\) after \(25.0 \mathrm{mL}\) of 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 \(\mathrm{pH}\) after \(300.0 \mathrm{mL}\) of HCl has been added. f. At what volume of HCl added does the \(\mathrm{pH}=6.04 ?\)

The reaction between HONH\(_2\) and HCl is: HONH\(_2\) + HCl → HONH\(_3^+\) + Cl\(^-\) Step 2: Calculating moles of HONH\(_2\)

We have 100.0 mL of 0.200 M HONH\(_2\). Therefore, the moles of HONH\(_2\) are: Moles of HONH\(_2\) = (Volume × Concentration) = (100.0 mL × 0.200 M) = 0.0200 mol HONH\(_2\) Step 3: Calculating moles of HCl at each stage

Use the volume of HCl that was added at each stage of the titration to calculate the moles of HCl: Moles of HCl = (Volume × Concentration) a. 0.0 mL of 0.100 M HCl = 0.000 mol HCl b. 25.0 mL of 0.100 M HCl = 0.00250 mol HCl c. 70.0 mL of 0.100 M HCl = 0.00700 mol HCl d. At the equivalence point, equal moles of HONH\(_2\) and HCl will have reacted, so 0.0200 mol HCl is added. e. 300.0 mL of 0.100 M HCl = 0.0300 mol HCl Step 4: Determine species present for each stage and calculate pH

For each stage, calculate the remaining moles of HONH\(_2\), HCl, and the formed HONH\(_3^+\), and use the Kb value (1.1 × 10\(^{-8}\)) to find the pH. a. 0.0 mL HCl added, buffer solution: HONH\(_2\) and HONH\(_3^+\) (with 0 added HCl) will be present, use Kb value to calculate pH. Consider HONH\(_2\) as the base and HCl as the acid. pH = 14 - pOH pOH = -log[\(OH^-\)] Kb = [HONH\(_3^+\)][OH\(^-\)] / [HONH\(_2\)] 1.1 × 10\(^{-8}\) = [\(x\)][\(x\)] / [0.200 - \(x\)] b. 25.0 mL HCl added, buffer solution: 0.00250 mol HCl reacts with HONH\(_2\), and we have excess HONH\(_2\). Use Kb value, the moles of HONH\(_2\), and HONH\(_3^+\) to calculate pH. c. 70.0 mL HCl added, buffer solution: 0.00700 mol HCl reacts with HONH\(_2\), and we have excess HONH\(_2\). Use Kb value, the moles of HONH\(_2\), and HONH\(_3^+\) to calculate pH. d. Equivalence point, 0.0200 mol HCl added: We have only HONH\(_3^+\) (a weak acid) and Cl\(^-\). Use the Kb value and water dissociation constant, Kw (1.0 × 10\(^{-14}\)), to calculate Ka for HONH\(_3^+\): Ka = Kw / Kb Calculate the concentration of HONH\(_3^+\) in the solution and use the Ka value to find the pH. e. 300.0 mL HCl added, acidic solution: Excess HCl is present, use the remaining moles of HCl and total volume to find the H\(^+\) concentration and calculate the pH directly. f. pH = 6.04, find volume of HCl added: Set the pH and use Kb value and moles of HONH\(_2\) and HONH\(_3^+\) to find the moles of HCl added. Then, use the concentration of HCl to find the volume of HCl added.