In a titration, 10.0 ml of 0.0750 M HCl(aq) is exactly neutralized by 30.0 ml of KOH(aq) of unknown concentration. What is the c
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Answer:
a. The maximum volume of 0.143 M HCl required is 154.4 mL.
b. The maximum volume of 0.143 M HCl required is 135.7 mL.
Explanation:
a.
Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)
Moles of aluminum hydroxide =
According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :
of HCl.
Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)
Moles of magnesium hydroxide =
According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :
of HCl.
Total moles of HCl required to neutralize both :
0.01346 mol + 0.008621 mol = 0.02208 mol
Molarity of the HCL solution = 0.143 M
Volume of the solution = V
1 L = 1000 mL
0.1544 L = 154.4 mL
The maximum volume of 0.143 M HCl required is 154.4 mL.
b.
Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)
Moles of calcium carbonate =
According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :
of HCl.
Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol
Molarity of the HCL solution = 0.143 M
Volume of the solution = V
1 L = 1000 mL
0.1357 L = 135.7 mL
The maximum volume of 0.143 M HCl required is 135.7 mL.