1. An unknown solid was added to a formic acid solution, which increased the pH
ID: 515731 • Letter: 1
Question
1. An unknown solid was added to a formic acid solution, which increased the pH of the solution. Is the unknown substance NaNO3 or NaCHO2? Explain how you arrived at your decision. Formic acid is a weak acid with a molecular formula of HCHO2. In aqueous solution it is in equilibrium with its dissociated and undissociated forms:
HCHO2(aq) H+(aq) + CHO2(aq)
2. Consider the conceptual reaction of …. A 2B
The forward reaction is endothermic. A is a blue substance, and B is a yellow substance. Together they make a green mixture. If this equilibrium mixture is heated up in a water bath, how will the color of the mixture change? Explain how you determined this.
Explanation / Answer
1)
Formic acid is a weak acid so there will be partial ionization to form some H+ and CHO2 ion, but the predominant species in solution is formic acid.
HCHO2 (aq) H+(aq) + CHO2(aq)
The pH will be below 7, which depends on the concentration of formic acid and the pKa of formic acid.
Sodium formate solution dissociate as following-
NaCHO2 (aq) --------->Na+(aq) + CHO2–(aq)
The sodium formate solution has a high concentration of sodium ions and formate ions and a small concentration of formic acid. The pH is above 7 and depends upon the concentration of sodium formate and the pKa of formic acid.
When we mixed sodium formate to formic acid solution, the formic acid solution adds sodium ions and formate ions. This disturbs the formic acid equilibrium, driving it towards reactants. This removes some hydronium ion from solution, increasing the pH of solution. Addition of the common ion to the formic acid equilibrium increases the PH.
NaNO3 is a neutral salt, so PH does not change, when it is added to formic acid.
Conclusion- The unknown substance is NaCHO2
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