To find pH for each equivalence point.
For HNO3 + NaOH ==> NaNO3 + H2O
That is a strong acid vs a strong base. Neither Na^+ nor NO3^- is hydrolyzed; therefore, the pH at the equivalence point is just that of pure H2O which is 7.0.
For NaOH + HCN ==> NaCN + H2O
The CN^- is a strong enough base to pull H^+ away from H2O; therefore, its pH is determined by the hydrolysis of th CN^-.
..........CN^- + HOH ==> HCN +OH^-
I........0.125............0....0
C.........-x..............x....x
E......0.125-x............x....x
Kb for CN^- = (Kw/Ka for HCN) = (x)(x)/(0.125-x). Solve for x = (OH^-) and convert to pH.
You can work this out but I obtained approx 11.
How did I arrive at 0.125M for CN^-? The titration of 0.25M HCN with 0.25M NaOH will require an equal volume of base added to whatever volume HCN was used initially so that dilutes th salt formed by a factor of 2.
Explain why methyl red would be a good indicator for a titration of 0.25M NaOH with .025M HNO3, but a poor choice for use in a titration of 0.25M NaOH with 0.25M HCN.
I'm pretty sure the first step is to find the pH of the two titrations (and then compare those to the pH range of methyl red), but I don't know (or can't remember) how..
2 answers
Thanks :)