À température ambiante: a) Calculer le pH d'une solution aqueuse \( 2,0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \mathrm{HCl} \) b) Calculer le pH d'une solution aqueuse \( 3,0 \times 10^{-8} \mathrm{~mol} / \mathrm{L} \mathrm{HCl} \) c) Calculer le pH d'une solution aqueuse \( 1,0 \times 10^{-9} \mathrm{~mol} / \mathrm{L} \mathrm{HCl} \)

Alright, I need to calculate the pH of three different hydrochloric acid (HCl) solutions at room temperature. The pH is a measure of the acidity or basicity of an aqueous solution and is defined as the negative logarithm (base 10) of the hydrogen ion concentration. The formula for pH is: \[ \text{pH} = -\log_{10}[\text{H}^+] \] Given that HCl is a strong acid, it dissociates completely in water, meaning that the concentration of hydrogen ions (\([\text{H}^+]\)) in the solution is equal to the concentration of HCl. This simplifies the calculation since I don't have to account for any undissociated HCl molecules. Let's tackle each part step by step. **Part a)** The concentration of HCl is \(2.0 \times 10^{-3} \, \text{mol/L}\). 1. **Identify the hydrogen ion concentration:** Since HCl is a strong acid, \([\text{H}^+] = 2.0 \times 10^{-3} \, \text{mol/L}\). 2. **Calculate the pH:** \[ \text{pH} = -\log_{10}(2.0 \times 10^{-3}) \] To compute this, I'll use the logarithm properties: \[ \log_{10}(2.0 \times 10^{-3}) = \log_{10}(2.0) + \log_{10}(10^{-3}) = 0.3010 - 3 = -2.6990 \] Therefore, \[ \text{pH} = -(-2.6990) = 2.6990 \] Rounding to two decimal places, the pH is approximately **2.70**. **Part b)** The concentration of HCl is \(3.0 \times 10^{-8} \, \text{mol/L}\). 1. **Identify the hydrogen ion concentration:** \([\text{H}^+] = 3.0 \times 10^{-8} \, \text{mol/L}\). 2. **Calculate the pH:** \[ \text{pH} = -\log_{10}(3.0 \times 10^{-8}) \] Breaking it down: \[ \log_{10}(3.0 \times 10^{-8}) = \log_{10}(3.0) + \log_{10}(10^{-8}) = 0.4771 - 8 = -7.5229 \] Therefore, \[ \text{pH} = -(-7.5229) = 7.5229 \] Rounding to two decimal places, the pH is approximately **7.52**. **Part c)** The concentration of HCl is \(1.0 \times 10^{-9} \, \text{mol/L}\). 1. **Identify the hydrogen ion concentration:** \([\text{H}^+] = 1.0 \times 10^{-9} \, \text{mol/L}\). 2. **Calculate the pH:** \[ \text{pH} = -\log_{10}(1.0 \times 10^{-9}) \] Simplifying: \[ \log_{10}(1.0 \times 10^{-9}) = \log_{10}(1.0) + \log_{10}(10^{-9}) = 0 - 9 = -9 \] Therefore, \[ \text{pH} = -(-9) = 9 \] The pH is exactly **9**. **Summary:** - **a)** pH = 2.70 - **b)** pH = 7.52 - **c)** pH = 9.00 These calculations show that as the concentration of HCl decreases, the pH increases, indicating a decrease in acidity and an increase in basicity. This makes sense because lower concentrations of HCl result in fewer hydrogen ions, leading to a less acidic solution.