{2.32 / [molar mass Al(OH)3(s)]} * 3 = moles (OH⎺)
each mole of OH⎺ needs one mole of HCl to neutralize it
Calculate the volume of 0.128 M HCl solution required to neutralize 2.32 g of Al(OH)3(s)
2 answers
The problem with this problem is the solubility of Al(OH)₃. With a Ksp = 4 x 10ˉ¹⁵, the solubility of Al(OH)₃ will be ~1.1 x 10ˉ⁴M in Al(OH)₃ dissolved (the rest will ppt) and the [OHˉ] = 3(1.1 x 10ˉ⁴)M = 3.3 x 10ˉ⁴ mol/L in OHˉ b/c Al(OH)₃ => Al⁺³ + 3OHˉ. So, the 2.32-g of Al(OH)₃ when added into 1 L solution will deliver only 3.3 x 10ˉ⁴ mole in OHˉ ions. If this is filtered and titrated, then...
(M x V) HCl = (M x V) OHˉ
(0.128M)(V-Liters) = (0.00033 mol/L OHˉ)(1 Liter)
V-Liters HCl needed = (0.00033 mol)/(0.128-mol/L) = 0.00234-L = 2.34-ml of 0.128M HCl needed.
NOTE => However, if both the ppt and solution are being titrated with HCl, then as the OH^- is converted to H2O on addition of HCl, more Al(OH)3 will dissolve into solution. Such will then require ~ 697-ml of 0.128M HCl to completely convert all the OH^- to water.
(M x V) HCl = (M x V) OHˉ
(0.128M)(V-Liters) = (0.00033 mol/L OHˉ)(1 Liter)
V-Liters HCl needed = (0.00033 mol)/(0.128-mol/L) = 0.00234-L = 2.34-ml of 0.128M HCl needed.
NOTE => However, if both the ppt and solution are being titrated with HCl, then as the OH^- is converted to H2O on addition of HCl, more Al(OH)3 will dissolve into solution. Such will then require ~ 697-ml of 0.128M HCl to completely convert all the OH^- to water.
1 answer