20cm3 of dilute HCl was found to dissolve 1.5g of CaCO3, calculate the molarity of the acid?

Question

To find the number of moles, do you either use the formula, moles=Mass/M(r)
or
moles=molarity x (volume in Litres)?
any advice would be helpful
thanks

Answer 1

first, we need to identify the pertinent reaction(s) involved here.

HCl ~> H+ + Cl-
CaCO3 ~> Ca2+ + CO32-

In order for the CaCO3 to dissolve, CaCl2 (as well as CaCl-) will need to be formed, and H2CO3 (as well as HCO3-) will need to be formed.

Whatever the Ksp value of CaCO3 is in HCl, I will not get into it, because your prolly not there yet - and this may cause confusion. As Ca and CO3 dissociate in the acidic HCl solution, Cl will find its way to Ca, and push the equilibrium so that the rest of the CaCO3 will dissolve. So the reaction that will help us is:

Ca2+ + 2Cl- ~> CaCl2 which is a salt, and which will dissolve in aqueous medium.

Lets do the math now.

[1.5g CaCO3 x (1 mol CaCO3 / ___ g CaCO3) x (1 mol Ca2+ / 1 mol CaCO3) x (2 mol Cl- / 1 mol Ca2+) x (1 mol HCl / 1 mol Cl-)] / .020L Solution = your answer (will be in Molarity)

Fill in the ____ with the molar mass of CaCO3, and perform the calculation of everything in [brackets] OVER the volume (.020L, which is = 20cm^3 = 20mL).

Answer 2

1.5g CaCO3 represent 1.5/100 moles of CaCO3.
Now:

CaCo3 + 2HCl -> CaCl2 + CO2 + H2O

So for each mole of CaCO3 we need 2 moles of HCl.
So if we have 1.5/100 Moles CaCO3 we must have 3/100 moles HCl = 0.03
These moles are present in 20 mls of dilute acid so in 1litre of this acid we must have 1000/20 times as much. This means we have 50*0.03 moles of HCl in 1 litre which works out to1.5 moles. Hence the HCl is 1.5 molar.

Answer 3

First find the moles of CaCO3.
1 x Ca = 1 x 40 = 40
1 x C = 1 x 12 = 12
3 x O = 3 x 16 = 48
40 + 12 + 48 = 100 (Mr CaCO3)
Mol(CaCO3) = 1.5/100 = 0.015 mol.
Next eq'n for reactants :-
CaCO3 + 2HCl = CaCl2 + CO2 + H2O
The molar ratios of the reactants are
CaCO3 : HCl :: 1 : 2
So one mole CaCO3 requires two moles HCl
So 0.015 mol CaCO3 requires 0.03 moles HCl.
Using moles = molarity x (volume/dm^3)
0.03 = [conc] x 20 cm^3/1000cm^3
[conc] = 0.03 mol x 1000 cm^3/ 20 cm^3
[conc] = 1.5 mol dm^3 = 1.5 M

Answer 4

22 drops of HCl = 1.000 mL so 1 drop of HCl = 0.045 mL; thus, the diluted HCl contains 15.000 mL + 1.000 mL + 0.045 mL = 16.045 mL. The 0.100 M HCl contains 0.100 mole HCl per liter so the 1.045 mL that were added contained 0.00010454 mole HCl. In the diluted solution that 0.00010454 mole HCl is in 16.045 mL so 0.00010454 mole HCl ÷ 16.045 mL * 1,000 mL per L = 0.0065154 M HCl. For strong acids like HCl the pH is the negative logarithm of the molarity calculated by adding 3 for thousandths to the logarithm of 6.5154. The logarithm of 6.52 is 0.8142476; the logarithm of 6.51 is 0.8135810 so 54 percent of the difference is 0.0003600; thus, the logarithm of 6.5154 is 0.0003600 + 0.8135810 = 0.8139410. The pH of 0.0065154 M HCl is 3.81394.