QUESTIONS Explain why you should use a sample veight that will use about 30 mL o
ID: 1037016 • Letter: Q
Question
QUESTIONS Explain why you should use a sample veight that will use about 30 mL of Na0H at the end point of the titration 1. 2. A 30.00 mL sample of 0.100MHOBr is titrated using 0.100 MNaOH Ka- 2.3110 for HOBr a) Write a balanced net-ionic equation for the titration reaction b) Calculate the pH of the titration mixture at the equivalence point. c) Would bromthymol blue be a suitable indicator for the titration? Refer to the discussion of indicators in your textbook Explain your answer. Why is it necessary to determine pH v.volume of titrant for volumes beyond the equivalence point in this experiment? 3. 4. Ifyour measured pKa value and equivalent veight values do not agree exactly with values in Table 1. vhich should you trust more. the pKa or the equivalent veight? Why?Explanation / Answer
Hello,
This question pertains to the behaviour of ions derived from acids and bases in an aqueous solution while they try to interact with each other. Many such ion interactions result in the establishment of equilibrium at various stages of titration and thus they result in an ionic equilibrium.
Since the experimental design isn’t mentioned in the question 1, for which 30 ml of NaOH is used so I am going to solve the rest of the parts. If you update the question in the comment section, I’ll be happy to help you with question 1 as well.
Q2. A 30 ml sample of 0.1 M HOBr is titrated using 0.1 M NaOH (Ka pf HOBr = 2.5 x 10-9)
a) write a balanced net ionic reaction for the titration reaction.
The balanced reaction can be arrived at by considering the number of H+ ions given out by HOBr and the number of OH- ions given out by NaOH, combining them and then rearranging the rest of the ions. In this case the balanced reaction would look like:
HOBr + NaOH = NaOBr + H2O
b) Calculate the pH of the titration mixture at the equivalence point.
The equivalence point of any titration refers to the stage at which the number of moles of H+ derived from the acid is equal to the number of moles of OH- derived from the base. At the equivalence point, thus the titration mixture contains only salt and water.
In the titration of HOBr and NaOH, we have a weak acid reacting with a strong base. HOBr behaves as a weak acid since the Ka value for it is quite small. Now, when this titration reaches the equivalence point, the mixture contains a salt of weak acid and strong base (NaOBr) in water. At this stage, two things happen:
i) ionization of NaBr to Na+ & OBr- in water
ii) reaction of Na+ & OBr- with water. This results in an equilibrium in and the process is known as hydrolysis. If we analyse the reaction, it would look like:
Na+ + H2O = NaOH + H+
OBr- + H2O = HOBr + OH-
Of these two reactions, the first one is less favourable than the second as it looks like the reformation reaction of NaOH for which dissociation is more favourable than reformation but the second one looks like the reformation reaction of weak acid HOBr, for which reformation using the ions is more favourable than ionization into respective ions.
In the second reaction, since OH- ions are liberated so the solution turns basic. The pH of this solution can be identified using the following method:
Millimoles of HOBr = 30 ml x 0.1 M = 3 mmoles
Millimoles of NaOH = 30 ml (required) x 0.1 M = 3 mmols [30 ml of NaOH is required since at equivalence point the moles of acid has to equal to base)
Millimoles of salt NaOBr = 3 mmoles
Total volume of solution = 30 ml (HOBr) + 30 ml (NaOH) = 60 ml
Concentration of salt NaOBr = 3 mmols/60 ml = 0.5 x 10-3 M
pH of the solution of salt of a strong base and weak acid = 7 + [pKa/2] + [log C/2] {C = concentration of salt}
i.e. pH of solution = 7 + [8.6/2] – [3.3/2] = 9.649
c) Would bromothymol blue be a suitable indicator for the titration?
When we need to choose an indicator, we have to keep two things in mind:
i) The equivalence point is when the colour changes most rapidly, not when the solution has changed colour.
ii) Indicators are weak organic acids or bases and are very sensitive to changes in pH. An indicator is characterised by its pKIn value, which is typically the pKa of the indicator. The most suitable indicator to mark the equivalence point of any titration is the one for which the pKIn value is the closest to the pH of equivalence point in this titration. Improper use of indicators will introduce inaccuracy to titration results.
The pKIn for bromothymol blue is 7.1 while the pH of the titration mixture at equivalence point is 9.64, so bromothymol blue won’t function as a suitable indicator for this titration. The pkIn for phenolphthalein is 9.3 and can function as a suitable indicator for this titration.
Q3. Why it is necessary to determine the pH vs volume of titrant for volumes beyond the equivalence point in this titration?
This is necessary because:
a) It helps us identify the end point of the titration graphically. It helps us observe the change in pH after end point, which more or less resembles a plateau. It suggests very less change in the pH of the solution beyond end point and helps us distinguish it from the end point, as bout the end point we’d observe the maximum change in pH.
b) adding a small amount of titrant after equivalence point helps giving a darker colour of the indicator, which helps us to distinguish it from the equivalence point at which colour changes abruptly
Q4. If your measured pKa value and equivalent weight values do not agree exactly with values in table 1, which should you trust more.
Table 1 is not represented in the question so I’d answer this question considering a generic titration. pKa measurement is temperature dependant while equivalent weight is temperature independent so equivalent weights would always give more accurate results as the experimental outcome wouldn’t depend on the temperature conditions.
Hope this helps. I request you to take some time, Rate the answer & drop in your valuable feedback.
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Thanks