a. Find the pH of orange juice if the concentration of hydrogen ion in the juice is \( \left[\mathrm{H}^{+}\right]=6.5 \times 10^{-4} \) b. Find the pH of milk if the concentration of hydrogen ion in the juice is \( \left[\mathrm{H}^{+}\right]=1.88 \times 10^{-6} \). c. Find the concentration of hydrogen ions in a glass of wine if the pH is 4.75 . d. Find the concentration of hydrogen ions in a bottle of vinegar if the pH is 5.75 .

**a. Finding the pH of orange juice** The pH is defined as \[ \mathrm{pH} = -\log \left[\mathrm{H}^{+}\right] \] Substitute \[ \left[\mathrm{H}^{+}\right] = 6.5 \times 10^{-4} \] Then, \[ \mathrm{pH} = -\log \left(6.5 \times 10^{-4}\right) = -\left(\log 6.5 + \log 10^{-4}\right) \] Using the property of logarithms, \[ \log 10^{-4} = -4 \] Thus, \[ \mathrm{pH} = -\left(\log 6.5 - 4\right) = 4 - \log 6.5 \] Approximating \[ \log 6.5 \approx 0.8129, \] we obtain, \[ \mathrm{pH} \approx 4 - 0.8129 = 3.1871 \] So, the pH of the orange juice is approximately \[ \boxed{3.19} \] --- **b. Finding the pH of milk** Given, \[ \left[\mathrm{H}^{+}\right] = 1.88 \times 10^{-6} \] Apply the pH formula: \[ \mathrm{pH} = -\log \left(1.88 \times 10^{-6}\right) = -\left(\log 1.88 + \log 10^{-6}\right) \] Since \[ \log 10^{-6} = -6, \] we have, \[ \mathrm{pH} = -\left(\log 1.88 - 6\right) = 6 - \log 1.88 \] Approximating \[ \log 1.88 \approx 0.2742, \] results in, \[ \mathrm{pH} \approx 6 - 0.2742 = 5.7258 \] Thus, the pH of the milk is approximately \[ \boxed{5.73} \] --- **c. Finding the concentration of hydrogen ions in a glass of wine** Given \[ \mathrm{pH} = 4.75 \] Rearrange the definition of pH: \[ \left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}} \] Then, \[ \left[\mathrm{H}^{+}\right] = 10^{-4.75} \] This evaluates to approximately, \[ \boxed{1.78 \times 10^{-5}\ \text{M}} \] --- **d. Finding the concentration of hydrogen ions in a bottle of vinegar** Given \[ \mathrm{pH} = 5.75 \] Again, using \[ \left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}, \] we have, \[ \left[\mathrm{H}^{+}\right] = 10^{-5.75} \] This evaluates to approximately, \[ \boxed{1.78 \times 10^{-6}\ \text{M}} \]