Final answer:
To determine the pH of a solution consisting of 10 CC of N/2 HCl, 40 CC of N/10 HNO3, and 60 CC of N/5 H2SO4, the moles of H+ from each acid are calculated and added together. The concentration of H+ is then used to calculate pH, resulting in an approximate pH of 0.52, which is strongly acidic.
Step-by-step explanation:
To calculate the pH of the solution where 10 CC of N/2 HCl, 40 CC of N/10 HNO3, and 60 CC of N/5 H2SO4 are mixed, we must first determine the concentration of hydrogen ions protons (H+) each solution contributes. Then, these values are summed together to find the total concentration of hydrogen ions in the solution.
- For HCl, which is a strong acid, the concentration of H+ is equal to its molarity. N/2 HCl is 0.5 M, so 10 CC (0.01 L) contributes 0.01 L * 0.5 mol/L = 0.005 moles of H+.
- For HNO3, also a strong acid, N/10 equates to 0.1 M. 40 CC (0.04 L) gives 0.04 L * 0.1 mol/L = 0.004 moles of H+.
- For H2SO4, which can donate two protons, N/5 is equal to 0.2 M. However, not all H2SO4 molecules will ionize completely. Let's assume complete ionization for this exercise (which is typically done in high school level problems), then 60 CC (0.06 L) of H2SO4 provides 0.06 L * 0.2 mol/L * 2 = 0.024 moles of H+.
The total volume of the solution is 10 CC + 40 CC + 60 CC = 110 CC or 0.11 L. The total moles of H+ are 0.005 moles + 0.004 moles + 0.024 moles = 0.033 moles.
The concentration of H+ is 0.033 moles / 0.11 L = 0.3 M. The pH is calculated using the formula pH = -log[H+]. The pH of this solution is therefore -log(0.3) approximately 0.52, indicating a strongly acidic solution.