Answer all the questions in the spaces provided.
- You are provided with:
- Solution X1 which is 0.1 M NaOH
- Solution X2 which is a dibasic acid ,H2Z
- Phenolphthalein indicator.
You are required to determine the concentration of the dibasic acid and the mass of the anion Z in the acid
Procedure
Fill the burette with solution X2 .pipette 25cm3 of solution X1 in to a conical flask and add two to three drops of phenolphthalein indicator .titrate X1 against X2 .Repeat this procedure two more times and complete the table below .
I II III Final burette reading cm3 Initial burette reading cm3 Volume of solution X2 used cm3 - Determine the average volume of solution X2 used. 2mks
- Workout the number of moles of NaOH solution X1 used. 3mks
- Given that the equation for the reaction is :
H2Z(aq) + 2NaOH(aq) → Na2Z(aq) + 2H2O(l)
Determine the number of moles of the dibasic acid H2Z used 3mks - Work out the concentration of solution X2 in moles per dm3. 3mks
- Given that the concentration of solution X2 is 8.17g/dm3 determine the relative formula mass of H2Z. 3mks
- Find the molar mass of the anion Z−2 in the acid H2Z(H= 1.0) 3mks
- You are provided with solid M .carry out the test below on M and fill the observation and inferences.
Test observation Inferences Observe solid M and describe its appearance 2mks 1mk Take half of solid M in a clean dry boiling tube and heat it gently 2mks 1mk Take the remaining solid M in to a boiling tube and add about 5cm3 of distilled water and shake .dived the solution formed in to two portions. 2mks 1mk To the first potion add a few drops of ammonia solution and then in excess 2mks 1mk To the second portion add a few drops of barium chloride followed by a few drops of dilute hydrochloric acid. 2mks 1mk
Hence write down the cation and anion in solid M .3mks
Marking Scheme
-
I II III Final burette reading cm3 15.0 14.9 15.0 Initial burette reading cm3 0.0 0.0 0.0 Volume of solution X2 used cm3 15.0 14.9 15.0
Consistency in the use of decimal places ✓1½
Accuracy ✓2 (Titre should be around 15.0cm3)- 15.0 + 14.9 = 14.95cm3 (correct principle of averaging)
2 - No. of moles of NaOH = 25 × 0.1 ✓1 = 0.0025moles ✓2
1000 - Mole ratio H2Z: NaOH = 1:2 ✓1
∴ moles of H2Z = 0.0025 ✓1 = 0.00125moles ✓1
2 - 14.95cm3 H2Z contains 0.00125moles
∴ 1000cm3 H2Z contains 1000 × 0.00125 =0.084M
14.95 - Molar Mass = concentration = 8.17 = 97
Molarity 0.084 - (1×2) + Z = 97
2+Z = 97
Z= 95
- 15.0 + 14.9 = 14.95cm3 (correct principle of averaging)
-
Test observation Inferences Observe solid M and describe its appearance Blue crystals M is likely to be a Copper (II) salt Take half of solid M in a clean dry boiling tube and heat it gently White powder formed colourless liquid formed M is hydrated salt Take the remaining solid M in to a boiling tube and add about 5cm3 of distilled water and shake .dived the solution formed in to two portions. Solid dissolves forming pale-blue solution M is likely to be a Copper (II) salt To the first potion add a few drops of ammonia solution and then in excess Pale-blue precipitate formed which dissolves in excess to form deep-blue solution. Cu2+ Present To the second portion add a few drops of barium chloride followed by a few drops of dilute hydrochloric acid. White precipitate formed insoluble in dil. HCl Cu2+ Present
Anions - SO42− ✓1½
Join our whatsapp group for latest updates
Tap Here to Download for 50/-
Get on WhatsApp for 50/-
Download Chemistry Paper 3 Questions and Answers - Form 3 End Term 1 Exams.
Tap Here to Download for 50/-
Get on WhatsApp for 50/-
Why download?
- ✔ To read offline at any time.
- ✔ To Print at your convenience
- ✔ Share Easily with Friends / Students