Question

Ph of 0.1 m nacn solution is 11 then the oercebtage hydrolysis is

Answer 1

Answer:

To find the pH of a 1.0 M solution of NaCNNaCN, given Ka(HCN)=4.9E(−10)Ka(HCN)=4.9E(−10)

The solution (refer end of question) uses three equations:

Equation 1 : HCN+H2O⟷CN−+H+HCN+HX2O⟷CNX−+HX+

Equation 2: NaCN⟷CN−+Na+NaCN⟷CNX−+NaX+

Equation 3: CN−+H2O⟷HCN+OH−CNX−+HX2O⟷HCN+OHX−

I am confused as to:

1) the logic behind why we need these three equations (up to this point in the course, all the similar questions have been of the form e.g. find pH of NH3 given Kb(NH3), and so we only used the NH3 acid base reaction equation.

I assume equation 1 is necessary because the Ka is given for HCN and Equation 2 is necessary because we are asked to find the pH of NaCN. Where does Equation 3 come from? We did talk about spectator ions in class so does it have something to do with Na being a spectator ion?

2) why the ICE (intial, change, at equillibrium) table to find x is written in terms of Equation 3 and not, for example, Equation 2.

The solution from the textbook is given belowTextbook Solution:

Textbook Solution

Thank you.

Answer 2

Answer:

When titrating the liberated iodine with the standardized thiosulfate titrant, the starch is only added near the end point to avoid decomposition due to the high iodine concentration. The analyst has a good indication that the end point is near as the solution color typically changes from a dark red to a light yellow.