Ch. 13 #97  [ENDORSED]

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Michael Lonsway 3O
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Ch. 13 #97

Postby Michael Lonsway 3O » Sat Dec 03, 2016 2:03 pm

How would I calculate the final PH for both friend's?
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Re: Ch. 13 #97  [ENDORSED]

Postby Chem_Mod » Sun Dec 04, 2016 10:39 am

For Friend #1: the reaction between CaCO3 with HCl is:

2 HCl + CaCO3 ---> CaCl2 + CO2 + H2O

The total mole of stomach acid (HCl) = 0.100 L × 0.10 M = 0.010 mole

Moles of CaCO3 in 2 tablets = 0.015 mole

Since 1 mole CaCO3 will neutralize 2 mole of HCl, 0.0150 mole CaCO3 will neutralize all of the HCl in the stomach. If assuming all CO2 will be escaped from stomach, the pH in the stomach will be 7.00 in the first person (in reality, the pH will be < 7 because some of the CO2 will remain in the stomach).

For friend #2: the reaction of MgO with HCl is:

MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O(l)

Total moles of MgO = = 2.98 × 10-2 mole

Since 1 mole MgO will neutralize 2 mole of HCl, 0.010 mole HCl will consume 0.00500 mole MgO.
Therefore, the moles of MgO left over are:

2.98 × 10-2 - 0.00500 = 0.0248 mole MgO

MgO will react with H2O producing OH- (100% dissociation):

MgO(s) + H2O (l) ---> Mg2+ (aq) + 2 OH(aq)

[OH-] = (2 × 0.0248 mole)/(3× 4.93 × 10-3 L + 0.100 L) = 0.432 M

Note: 1 US tsp = 4.93 mL;

pOH = 0.36; pH = 13. 64

The pH in friend #2’s stomach is much higher than that of friend #1.

Ask me in Chem 14B if this isn't detailed enough. :-)


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