7. Suppose 45.0 mL of 0.250 M HNO3 was mixed with 35.0 mL 0.600 M Ba(OH)2. Calculate the pH of the final solution.
To calculate the pH of the final solution, we need to determine the concentration of H+ ions in the solution. This can be done by using the stoichiometry of the reaction between HNO3 and Ba(OH)2.
First, let's write the balanced equation for the reaction:
2 HNO3 + Ba(OH)2 -> 2 H2O + Ba(NO3)2
From the balanced equation, we can see that 2 moles of HNO3 react with 1 mole of Ba(OH)2. Therefore, the number of moles of HNO3 can be calculated using the formula:
moles of HNO3 = volume (in L) x concentration (in M)
For the HNO3 solution, we have:
moles of HNO3 = 0.045 L x 0.250 M = 0.01125 moles
Similarly, for Ba(OH)2, we have:
moles of Ba(OH)2 = 0.035 L x 0.600 M = 0.021 moles
Since 2 moles of HNO3 react with 1 mole of Ba(OH)2, we have an excess of Ba(OH)2. Therefore, all the HNO3 will react completely, and the excess Ba(OH)2 will be left.
Now, we need to determine the concentration of H+ ions in the solution. For each mole of HNO3, 2 moles of H+ ions are produced. This means that the final concentration of H+ ions is:
[H+] = 2 x (moles of HNO3 / total volume of the solution in liters)
The total volume of the solution is the sum of the volumes of HNO3 and Ba(OH)2, which is 45.0 mL + 35.0 mL = 80.0 mL = 0.080 L.
[H+] = 2 x (0.01125 moles / 0.080 L) = 0.28125 M
Now, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.28125) ≈ 0.550
Therefore, the pH of the final solution is approximately 0.550.