LEAVING CERT HIGHER CHEMISTRYPREDICT PAPER JUNE 2019

ALL THE PREDICTED PAPERS ARE BASED ON A PATTERN FOR THAT PARTICULAR SUBJECT.THEY CONTAIN ENOUGH MATERIAL FOR A REVISION ALSO. THEY PREDICT THE TOPICS THAT MIGHT COME UP AND GIVE EXAMPLES OF PAST QUESTIONS ON THESE TOPICS ANSWERING THEM IN DETAIL.WHAT SHOULD YOU DO WITH THIS PREDICTED PAPER READ IT..... THEN COVER THE ANSWERS AND TRY TO DO THE QUESTION ON A BLANK PIECE OF PAPER IF YOU ARE SUCCESSFUL AWARD YOURSELF A SMIL.ING GREEN MAN... HIGHLIGHT HIM GREEN..IF YOU CANT DO IT AWARD YOURSELF A UNHAPPY RED MAN ….KEEP STUDYINGUNTIL MOST-OR ALL OF THEM ARE GREEN .DONT BOTHER WITH ANY OTHER MATERIAL UNTIL THIS IS DONE. . IF THERE ARE ANY MISTAKES, WE DID NOT MEAN MAKE THEM.GOOD LUCK IN THE EXAMS

QUESTION 1 THE INORGANIC EXPERIMENT.NO 1 The water of crystalization in washing soda.

A sample of the crystals Na 2 CO 3 . xH 2 O was accurately weighed and found to have a mass of 2.50 g. The sample was dissolved in deionised water and made up to 250 cm3 of solution. A number of 25.0 cm3 portions of this solution were titrated with a previously standardised 0.10 M hydrochloric acid (Hcl) solution. The mean volume of the hydrochloric acid solution required to reach the end point was 21.6 cm3 The balanced equation for the titration reaction is: 2HCl + Na 2 CO 3 →2NaCl + H 2 O + CO 2

Explain the underlined term. Answer. The concentration or the molarity is exactly known

Describe in detail how the chemist should have dissolved the weighed sample of washing soda crystals and made thesolution up to exactly 250 cm3Answer.wash into a beaker with deionised water and stir to dissolve.When making up the solution pour through funnel into a volumetric flask adding the rinsings of beaker add the last few drops of deionised water drop by drop

so that the bottom of the meniscus is level with the mark on the flask.State the procedure near end point of the titration.

add drop at a time wash down inner sides of conical flask and agitate after each drop looking for the colour change.

Explain how this precaution would have contributed to theaccuracy of the titration result.

add dropwise so that end point detected as one drop of solution would change colour near end point wash sides so that all reagent in the reaction mixture reacts with other reagent and finally agitate so that reactants meet each other and react.

Name a suitable indicator for this titration.State the colour change in the titration flask at the end point.

methyl orange ….......yellow to pink From the mean volume of the hydrochloric acid solution, calculate the concentration of sodium carbonate

(Na 2 CO 3 ) in the original solution in (i) moles per litre, Answer

Molarity means moles per litre.

ii) grams per litre.Multiply the molarity by the molar mass of the sodiumcarbonate.

= .0432 X 106 = 4.6 grams per litre.

Calculate the percentage water of crystallisation in the crystals and the value of x, the average number of water molecules in the formula Na CO . xH O.

Hydrated water present at the start = 2.50 g/250 cm3 same as 10 grams per litre . This included the water...ie X by 4 both sides.

From the titration result there is 4.6 grams of the pure sodium carbonate...therefore amount of water present is

10 – 4.6 = 5.4 grams water.Therefore % water present in the 10 gram sample was 5.4/ 10 X 100 = 54%

find the ratio of water molecules to carbonate ones.Water : sodium carbonate5.4/18 : 4.6/106

.3 : .0433 now divide by smaller one

.3/.0433 and .o433/.o433

7 : 1 7 water to 1 carbonate.

NO 2 TO DETERMINE THE % HYPOCHLORITE IN BLEACHIntroduction.Bleaches are simply solutions of hypochlorite salts such as sodium hypochlorite (NaOCl) or calcium hypochlorite (Ca(OCl)2 ). In titrations the Hypochlorite ion reacts with excess iodide ion in the presence of acid to generate an iodine solution: ClO- + 2I- + 2H+ → Cl- + I2 + H2O The liberated iodine solution can be titrated against sodium thiosulfate solution using a freshly prepared starch solution as indicator. The titration reaction may be represented by the equation: I2 + 2S2O3 -2 → 2I- + S4O6 -2