Answer:
A. 9.83g
B. 13.06
Step-by-step explanation:
A) To calculate the mass of BaSO4 formed, you need to first write the balanced equation for the reaction:
Ba(OH)2 + H2SO4 -> BaSO4 + 2H2O
Then, you need to find the limiting reactant, which is the one that runs out first and determines how much product is formed. You can do this by converting the volumes and concentrations of the solutions to moles and comparing them with the stoichiometric coefficients.
50 mL of 1.00 M Ba(OH)2 = 0.050 L x 1.00 mol/L = 0.050 mol Ba(OH)2 88.7 mL of 0.475 M H2SO4 = 0.0887 L x 0.475 mol/L = 0.0421 mol H2SO4
According to the equation, 1 mol of Ba(OH)2 reacts with 1 mol of H2SO4, so Ba(OH)2 is in excess and H2SO4 is the limiting reactant.
Next, you need to use the mole ratio between the limiting reactant and the product to find how many moles of BaSO4 are formed:
0.0421 mol H2SO4 x (1 mol BaSO4 / 1 mol H2SO4) = 0.0421 mol BaSO4
Finally, you need to multiply the moles of BaSO4 by its molar mass to get its mass:
0.0421 mol BaSO4 x 233.39 g/mol = 9.83 g BaSO4
So, the mass of BaSO4 formed is 9.83 g.
B) To calculate the pH of the mixed solution, you need to first find the concentration of OH- ions that remain after the reaction. You can do this by subtracting the moles of OH- that reacted with H+ from the initial moles of OH- and dividing by the total volume of the solution.
The initial moles of OH- are equal to the moles of Ba(OH)2:
0.050 mol Ba(OH)2 x (2 mol OH- / 1 mol Ba(OH)2) = 0.100 mol OH-
The moles of OH- that reacted with H+ are equal to the moles of H2SO4:
0.0421 mol H2SO4 x (2 mol H+ / 1 mol H2SO4) = 0.0842 mol H+
The remaining moles of OH- are:
0.100 mol OH- - 0.0842 mol H+ = 0.0158 mol OH-
The total volume of the solution is:
50 mL + 88.7 mL = 138.7 mL = 0.1387 L
The concentration of OH- is:
0.0158 mol OH- / 0.1387 L = 0.114 M
Next, you need to use the relationship between pH and pOH to find the pH:
pOH = -log[OH-] = -log(0.114) = 0.94 pH + pOH = 14 pH = 14 - pOH = 14 - 0.94 = 13.06
So, the pH of the mixed solution is 13.06.