Consider the titration of 80.0 mL Ba(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.
a.) 0.0 mL
*Which molecular equation do I use?
b.) 20.0mL
Ba(OH)2__+2HCl_+=>+_BaCl2_+____2HOH
80.0 mL__20.0mL
0.100M___0.400M
.008 mol_.008 mol
.004 left_ 0 left__.004 mol produced
pH = pKa
pH = -Log(1.05x10^-12)
pH = 11.98
*Did I do this one correctly?
You didn't include the molarity of the Ba(OH)2.
0.100 M
You have correctly balanced the molecular equation for the reaction between Ba(OH)2 and HCl. However, the calculations for the moles of reactants and products are incorrect.
To calculate the moles of Ba(OH)2 and HCl, you can use the equation:
moles = concentration (M) × volume (L)
For the reaction at 0.0 mL of HCl added:
moles of Ba(OH)2 = 0.100 M × 0.080 L = 0.008 mol
moles of HCl = 0.400 M × 0.000 L = 0 mol
Since no HCl has been added, there are no moles of HCl reacting yet.
For the pH calculation, it seems like you are trying to use the pKa value of water (1.05 × 10^-12) to find the pH. However, in this reaction, water is not the only product. The reaction produces BaCl2 and H2O.
To find the pH, you need to consider the concentration of the hydroxide ions (OH-) in the solution resulting from the reaction. Since no HCl has been added, the concentration of OH- ions should remain the same as the initial concentration from Ba(OH)2. You can then use the concentration of OH- to find the pOH and convert it to pH.
So, in this case, you need to find the concentration of OH- ions in the solution at 0.0 mL of HCl added. In a neutral solution, the concentration of OH- ions is equal to the concentration of H+ ions, which can be calculated using the initial concentration of Ba(OH)2:
[OH-] = 2 × 0.100 M = 0.200 M
pOH = -log[OH-] = -log(0.200) ≈ 0.70
pH = 14 - pOH = 14 - 0.70 = 13.30
So, the pH of the resulting solution after adding 0.0 mL of HCl is approximately 13.30.
Please let me know if you have any further questions.
To calculate the pH of the resulting solution, you need to consider the balanced chemical equation for the reaction between Ba(OH)2 and HCl. The equation is:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
For part a) where 0.0 mL of HCl is added, no reaction occurs because there is no HCl to react with Ba(OH)2. Therefore, the concentration of OH- remains unchanged. To calculate the pH, you can use the equation pH = -log[OH-].
However, you haven't provided the concentration of Ba(OH)2 in this question, so it's not possible to calculate the pH using this information alone. Please provide the concentration of Ba(OH)2 to continue solving the problem.