If 27.5 mL of 5.7 multiplied by 10-2 M HNO3 is added to 20.0 mL of 2.2 multiplied by 10-2 M HNO3, what is the pH of the solution?
If 12.5 mL of 8.6 multiplied by 10-2 M NaOH is added to 32.5 mL of 6.0 multiplied by 10-4 M NaOH, what is the pH of the solution?
moles first HNO3 = M x L = ??
moles second HNO3 = M x L = ??
molarity of final solution = total moles/total volume.
Since HNO3 is a strong acid (100% ionized) the pH = -log(H^+) and (H^+) (HNO3)
To determine the pH of the solution, we need to calculate the concentration of H+ ions in the resulting solution after mixing the two solutions together. We can start by calculating the moles of H+ ions in both solutions.
For the first solution:
Volume = 27.5 mL = 0.0275 L
Concentration = 5.7 × 10^(-2) M
Moles = Concentration × Volume = (5.7 × 10^(-2)) × 0.0275 = 1.5675 × 10^(-3) moles
For the second solution:
Volume = 20.0 mL = 0.0200 L
Concentration = 2.2 × 10^(-2) M
Moles = Concentration × Volume = (2.2 × 10^(-2)) × 0.0200 = 4.4 × 10^(-4) moles
Now, we can add the moles of H+ ions from both solutions together:
Total moles of H+ ions = 1.5675 × 10^(-3) + 4.4 × 10^(-4) = 2.0075 × 10^(-3) moles
To find the total volume of the solution, we add the volumes of the two solutions:
Total volume = 27.5 mL + 20.0 mL = 47.5 mL = 0.0475 L
Now we can calculate the concentration of the H+ ions in the resulting solution:
Concentration of H+ ions = Total moles / Total volume = (2.0075 × 10^(-3)) / 0.0475 = 0.0422 M
Finally, we can find the pH of the solution using the formula: pH = -log[H+]
pH = -log(0.0422) ≈ 1.373