Antacid Suspension Question (1 Viewer)

[LankanG]

An antacid suspension bottle label statesl -
" Each 10ml contains 400mg (0.0069mol) of solid magnesium hydroxide and 400 mg (0.0051mol) of solid aluminium hyroxide, suspended in water"

People take antacid suspensions if they have indigestion from an excess of stomach acids. The acid in the stomach is HCl and has a pH of 2.0.

What volume of stomach HCl would be neutralised by 10ml of the antacid supension?

Thanks in advance
Cheers

 

[Vimal.K]

Ummmm. I see where this is going. This is one of those " If a tree falls in the forest, do the other trees laugh at it?" kinda questions right?

 

[brenton1987]

pH 2 = 0.01 mol.L^-1

---

2HCl + Mg(OH)₂ --> 2H₂O + MgCl₂

0.0069 mol Mg(OH)₂ reacts with 0.0138 mol HCl
0.0138 mol / 0.01 mol.L^-1
1.38 L

---

3HCl + Al(OH)₃ --> 3H₂O + AlCl₃

0.0051 mol Al(OH)₃ reacts with 0.0153 mol HCl
0.0153 mol / 0.01 mol.L^-1
1.53 L

---

Total acid neutralised is 2.91 L which seems extremely excessive.

 

[wrxsti]

why u treat them as 2 different values? doesnt that mean that your neutrilising them twice???

For the Magnesium tablet (full neutrilisation)
then the Aluminium tablet......

you get what im saying ? :S

 

[brenton1987]

you get what im saying ? :S

No.

 

[Forbidden.]

I think they are treated as two separate calculations because there are two compounds to be neutralised ?

 

[wrxsti]

No.

righteo.........
lol... cant you see that youve neutrilised the tablet twice?

 

[brenton1987]

righteo.........
lol... cant you see that youve neutrilised the tablet twice?

If 10 mL contains 0.0069 mol of Mg(OH)₂ and 0.0051 mol of Al(OH)₃ there is a total of 0.0291 mol of OH^- (0.0069 + 0.0069 + 0.0051 + 0.0051 + 0.0051)
0.0291 mol / 0.01 mol^.L^-1
2.91 L

Nothing has been neutralised twice.

Show me how you would work it out.