Question

Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutralize. Assume complete neutralization.
a. A tablet containting 350 mg Al(OH)3 and 250 mg Mg(OH)2.
b. A tablet containing 970 mg of CaCO3.

Answer 1

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.

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ 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

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ 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

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.