If I dissolve a Tums tablet(1000 mg of CaCO3) in 25 ml of water and then add 25 ml of a 0.1N HCl, what Is the pH of the mixture?

Question

I know that CaCO3 isn’t very soluble in water. I crushed the tablet and added water to create a suspension. I then measured the pH and found it to be on average about 9. If I then add 25 ml of 0.1 N HCl, I assume the acid will be completely neutralized leaving me with a pH also around 9. If this is correct, I would need to know how to mathematically calculate the original pH of CaCO3 in water and then the final pH after HCl is added.

Answer 1

If you were to completely neutralize the original solution, the pH would not be 9, but it would be closer to 7 (neutral pH). The 1000 mg unfortunately doesn't tell you how much CaCO₃ is in the tablet because a large majority of that mass is binder and inert ingredients.  Thus, there really is no way to know exactly how many mg of CaCO₃ is in the solution and thus you cannot find the number of moles of CaCO₃. Your question asks to calculate original pH of the CaCO₃.  You can't calculate it, but you already measured it at pH 9.

 

At any rate, if you have a pH = 9, that tells you that the [H⁺] is 1x10^-9 M.  Since you have 25 ml (0.025 L) of that solution, you have (0.025 L)(1x10^-9 mol/L) = 2.5x10^-11 moles H⁺. To this you have added 25 ml (0.025 L) of 0.1 N HCl (0.1 moles/L) so you have added 0.0025 moles H⁺ and now the final volume is 50 ml (0.05 L).

 

Normally, you would add the H⁺ in the original solution (2.5x10^-11 moles) and the H⁺ in the added HCl (2.5x10^-3 moles) and then divide by 0.05 L to get [H⁺] and then pH can be calculated from that.  However, since the amount of H⁺ in the original sample is so small relative to that which is added, we can neglect it, and the pH can be calculated as 

2.5x10^-3 moles H⁺/0.05 L = 0.05 M H⁺.  The pH of the final solution would then be -log 0.05 = 1.3.