[rosin]

Hi!

I'm a student from Europe, I don't speak English fluently, I'm sorry if the language is bad. I would appreciate it if anyone could help me with this task, I'm not so good in chemistry:

In a lake we can add calciumcarbonate

CaCO3 (s) + 2H+ (aq) + SO42- (aq) ⇌ Ca2+ (aq) SO42- + H2O (aq) + CO2 (aq)
 
CaCO3(s) + 2H+(aq)  <--> Ca2+(aq) + CO2(g) + H2O(l)

1 dm^3 of the water has a pH 4,6
Find out how much calciumcarbonate you have to add to the water to increase the pH to 5,60.

My equations so far
(10^-4,6 mol/L) * y = (10^-5,6 mol/L)
y = 0,1
In other words an increase with 10%

1 mole water gives 2 mole hydrogenions

I was thinking about finding the concentration of calsiumcarbonate when pH is 4,6 and then to multiplicate with 0,1.

Can somebody please help me?

Friendly regards

[Borek]

This is simple stoichiometry.

You know starting concentration of H⁺, you know final concentration. Calculate how much of H⁺ must be neutralized.

Then, assume the reaction:

CaCO3(s) + 2H+(aq)  <--> Ca2+(aq) + CO2(g) + H2O(l)

and just calculate number of moles of CaCO₃ required.

[rosin]

Thank you for your reply. Should I see 2H+ as the pH or H+ as the pH given?

Ok, this is my try:
10^-5,6 + H3O+ = 10^-4,2
H3O+ = 2,26 * 10^-5 mol/L (must be removed)

in this task the concentration of H3O+= mole H30+

The relation between CaCO3 and H3O+ = 1:2

The mole CaCO3 = (1/2) = 1/2 (mole H30+) = 1,13 * 10^-5

mole * molar mass = mass CaCO3 needed

1,13 *10^-5 mole * 100,19 g/mol =0,0011324 gram

Is this correct?

[Borek]

 

[rosin]

thank you so much!!:)

[rosin]

but if the reaction is

CO32- (aq) + H3O+ (aq) ⇌ HCO3- (aq) + H2O (l)

then the relation would be one to one? it gives me a slightly different answer