Consider the titration of 80.0 mL of 0.100 M Sr(OH)2 by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.
(b) 20.0 mL
(c) 30.0 mL
(d) 40.0 mL
(e) 80.0 mL
To calculate the pH of the resulting solution, we need to determine the amount of excess HCl or OH- ions present after each addition of HCl.
(b) 20.0 mL HCl added:
The initial moles of Sr(OH)2 present in the 80.0 mL solution can be calculated using the formula:
moles of Sr(OH)2 = volume (L) × concentration (M)
= 0.080 L × 0.100 M
= 0.008 mol
Since Sr(OH)2 is a strong base, it will react with the strong acid HCl in a 1:2 mole ratio. So, 0.008 mol of Sr(OH)2 will react with half the number of moles of HCl, which is 0.008 mol × 0.400 M / 2 = 0.0016 mol.
The volume of HCl added is 20.0 mL, which is equal to 0.020 L. The remaining moles of HCl after reaction can be calculated as:
moles of HCl = volume (L) × concentration (M)
= 0.020 L × 0.400 M
= 0.008 mol
The excess moles of HCl can be calculated as the initial moles of HCl, which is 0.008 mol, minus the moles of HCl that reacted with Sr(OH)2, which is 0.0016 mol. Therefore, the excess moles of HCl is 0.008 mol - 0.0016 mol = 0.0064 mol.
To calculate the concentration of the excess HCl, we divide the excess moles by the new volume of the solution, which is 80.0 mL + 20.0 mL = 100.0 mL = 0.100 L:
concentration of HCl = moles of HCl / volume (L)
= 0.0064 mol / 0.100 L
= 0.064 M
Since HCl is a strong acid, it is completely dissociated and contributes to the concentration of H+ ions in the solution. Therefore, the concentration of H+ ions is 0.064 M.
To calculate the pH, we use the formula:
pH = -log[H+]
= -log(0.064)
= 1.192