Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(8.3 \times 10^{-4} \mathrm{MHCl},(\mathbf{b}) 1.20 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(500 \mathrm{~mL}\) of solution, $(\mathbf{c}) 2.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{HClO}_{4}\( diluted to \)40.0 \mathrm{~mL}\(, (d) a solution formed by mixing \)25.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HBr}\( with \)25.0 \mathrm{~mL}\( of \)0.200 \mathrm{M} \mathrm{HCl}$.

The strong acid HCl fully dissociates in water, so the concentration of H⁺ ions is the same as the concentration of the solution: [H⁺] = 8.3 × 10⁻⁴ M. Use the pH formula to calculate the pH: \[\mathrm{pH} = -\log([\ce{H+}]) = -\log(8.3 \times 10^{-4})\] Now, use a calculator to find the pH value.

First, we need to find the concentration of the HNO₃ solution. The molar mass of HNO₃ is approximately 63.01 g/mol, so 1.20 g of HNO₃ is: \(\frac{1.20\,\text{g}}{63.01\,\text{g/mol}} \approx 0.019\,\text{mol}\) of HNO₃ Next, we calculate the molarity based on the volume: \[0.019\, \text{mol} \div 0.5\, \text{L} \approx 0.038\,\text{M}\] Since HNO₃ is a strong acid, the concentration of H⁺ ions is the same as the concentration of the solution: [H⁺] = 0.038 M. Apply the pH formula: \[\mathrm{pH} = -\log([\ce{H+}]) = -\log(0.038)\] Now, use a calculator to find the pH value.

First, we need to find the concentration of the diluted solution. We can use the dilution formula: \(M_1 V_1 = M_2 V_2\), where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. \[0.250\, \text{M} \times 0.002\, \text{L} = M_2 \times 0.040\, \text{L}\] Solving for M2: \[M_2 = \frac{0.250\, \text{M} \times 0.002\, \text{L}}{0.040\, \text{L}} \approx 0.0125\, \text{M}\] The concentration of H⁺ ions is the same as the concentration of the solution: [H⁺] = 0.0125 M. Apply the pH formula: \[\mathrm{pH} = -\log([\ce{H+}]) = -\log(0.0125)\] Now, use a calculator to find the pH value.

First, we need to find the total amount of H⁺ ions in the mixed solution. Since H⁺ ions come from both HBr and HCl, we need to calculate the moles of H⁺ ions from each acid: HBr: \(0.100\, \text{M} \times 0.025\, \text{L} = 0.0025\, \text{mol}\) HCl: \(0.200\, \text{M} \times 0.025\, \text{L} = 0.0050\, \text{mol}\) Now, add the moles of H⁺ ions to get the total moles of H⁺ ions: \[0.0025\, \text{mol} + 0.0050\, \text{mol} = 0.0075\, \text{mol}\] Next, we need the new total volume: \[0.025\,\text{L} + 0.025\,\text{L} = 0.050\,\text{L}\] Now, we can find the concentration of H⁺ ions in the mixed solution: \[\frac{0.0075\, \text{mol}}{0.050\, \text{L}} = 0.15\, \text{M}\] Finally, apply the pH formula: \[\mathrm{pH} = -\log([\ce{H+}]) = -\log(0.15)\] Now, use a calculator to find the pH value.