Question

Calculate the mass (mg) of Mg(OH)2 needed in order to neutralize stomach contents with a ph of 1.3 and a solution volume of 250ml. Assume stomach contents to be HCL solution. Start with a balanced reaction.

Answer 1

Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O

use:

pH = -log [H+]

1.3 = -log [H+]

[H+] = 5.012*10^-2 M

mol of H+ present = [H+]*volume of acid in L

= 5.012*10^-2 M * 0.250 L

= 0.01253 mol

for complete neutralisation,

mol of OH- required = mol of H+ reacted

= 0.01253 mol

mol of Mg(OH)2 = (1/2)*mol of OH-

= 0.01253 mol / 2

= 6.265*10^-3 mol

Molar mass of Mg(OH)2,

MM = 1*MM(Mg) + 2*MM(O) + 2*MM(H)

= 1*24.31 + 2*16.0 + 2*1.008

= 58.326 g/mol

use:

mass of Mg(OH)2,

m = number of mol * molar mass

= 6.265*10^-3 mol * 58.33 g/mol

= 0.3654 g

= 365 mg

Answer: 365 mg