The titration graph can be divided into the ten segments indicated below. For ea

Question

The titration graph can be divided into the ten segments indicated below. For each of these segments: Discuss the relative concentrations of H_2 A, HA^-, and A^2-. In any given part of the graph, one of these species can be dominant, or two can be of comparable concentration with the remaining species at a negligible concentration. Say which of the three is dominant, or which two are comparable, and which is/are negligible in each of the ten segments. Discuss the relative concentrations of H_3 O^+ and OH^-. Which of these is present at a higher concentration? If the segment is a region (indicated by the word "between" or "after"), discuss how the concentrations of the five species involved (H_2 A, HA^-, and A^2-, H_3 O^+ and OH^-) are changing as the titration proceeds across this region. The ten segments of the graph are:

Explanation / Answer

H2A------- HA-+H3O+ 1

HA- ------- A2- + H3O+ 2

At stage 1 there is higher concentration of H2A.But there is very low concentration of HA- and A2-. due to the little amount of disscociation

At stage 2, when the base is added to the H2A solution equilibrium point of the first reaction shifts to the right side and start to arise the HA- and H3O+ concentrations.

At stage 3,more and more H2A molecules started to breakdown into the H3O+ and HA-.HA concentration started to getting low at this point.

At stage 4,H2A concentration is very low and HA- and H3O+ is high at this end.sytem came to it's first equivalance point.

At stage 5,it passed the first equivalence point and all the H2A converted to HA- .pH change occured at the equvalence point and HA- started to dissociate in the system.

At the stage 6 when adding base to the system remaining HA- started to produce H3O+ and A2- and equillibrium point of second reaction shifted to the right hand side.

At stage 7,HA- concentration was getting very low and A2- and H3O+ concentration was getting high at this point.

At stage 8,HA- concentration was very low and A2- and H3O+ concentration is very high at this point.It's very close the 2 nd equivalence point.

At stage 9, all the HA- converted to the A2- and H3O+ at this point.It's 2 nd equivalence point.Highest H3O+ could be observed at this point

At stage 10, A2- remained in the solutiont.all the H3O+ is over at the point.remaining OH- of NaOH existed in the solution.

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