Chemistry, 05.10.2019 19:20 plug30

What volume of 2m hydrochloric acid will neutralise 25 cm3 sodium carbonate?

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In sodium carbonate I believe the carbonate is considered the base and that the HCl will react with the carbonate to produce bicarbonate (HCO₃). Due to that, I think the chemical equation should be HCl+NaCO₃→HCO₃+NaCl.

To neutralize the solution, you need to have equal amounts H⁺ and CO₃⁻ which means that you need to have equal amounts HCl and NaCO₃. To find this you need to know how many moles of NaCO₃ there are in the 25mL solution since that is the number of moles of HCl you need. You would then divide the number of moles of HCl needed by 2M to find the volume of HCl needed.

With the information given I don't think you can answer this. You need to know what the concentration of the sodium carbonate is and with out that value you can't determine how many moles of sodium carbonate there are. You need to know how many moles of sodium carbonate you have in order to know how much 2M hydrochloric acid is needed to neutralize it.

The process you would usually use is:
(0.025L NaCO₃)×(molarity of NaCO₃)=moles NaCO₃
moles NaCO₃=moles HCl needed to neutralize
(moles HCl needed to neutralize)/2M=Liters of HCl needed to neutralize

Let me know if the question came with any more information but right now I don't think it can be solved.

I hope this helps. Let me know if anything is unclear.

Also, since this reaction is between a strong acid and weak base, when you neutralize it with the strong acid the pH of the solution will not be 7. It will most likely be under seven since HCO₃ is being produced which is the conjugate acid of CO₃ and therefore create a slightly acidic solution at the equivalence point (the equivalence point is another way of saying the solution was neutralized). You probably don't need to know this for the scope of your class but I thought it was worth mentioning since later on you will have to deal with this fact.

Answer from: Quest

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Answer from: Quest

It shows that the experiment is designed properly.

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