Question

Consider the following four titrations: i. 100.0 mL of 0.10 M HCl titrated with 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated with 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated with 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated with 0.10 M NaOH. Rank the titrations in order of increasing pH at the equivalence point.

Answer 1

Answer:

Solution III < Solution II < Solution I < Solution IV

Explanation:

I) $HCl + NaOH \longrightarrow H_2O + NaCl$

Due to the rate of neutralization is 1 mol of NaOH with 1 mol of HCl and that we have equal volumes of both solution with the same concentration, they neutralize each other leaving a neutral pH (pH=7). Althought, because your titrating agent is a base the equivalence point will be a bit alkaline.  

II) The same as the point I) but now the titrating agent is the acid, the equivalence point will be neutral trending a bit to acid.  

III) Given that we have one strong acid neutralizing a weak base solution, in the equivalence point the the pH will be acid because the HCl (strong) dissociates compleately so we will have the 0.01 mol (100 ml of 0.1 M) of HCl with 0.01 mol>CH3NH2 (because is a weak base). How much less than 0.01 mol we have? It depends of the strengh of the base and the pH will also depend on that.

IV) This case is similar to the last one but with a strong base and a weak acid. So with a similar deduction the pH in the equivalence point will be alkaline (pH>7). How much? It depends on the strengh of the acid.

In order of increasing pH:

Solution III < Solution II < Solution I < Solution IV