# 2012 NECO CHEMISTRY PRACTICALS

Filed in Uncategorized by on June 12, 2012

VOLUMETRIC ANALYSIS:

1. A is 0.125mol/dm3 of H2SO4.

B is a solution containing xgdm^-3 of NaOH.

(a.) Put A into the burette and titrate it against 20.0cm3 or 25.0cm3 portion of B using Methyl Orange as indicator. Record the volume of your pipette.

Tabulate your burette readings and Calculate the average volume of A used.

THE EQUATION FOR THE REACTION IS:

H2SO4(aq) + 2NaOH –>Na2SO4 + 2H2O

From Your result and Information provided above, Calculate

(i) the amount of H2SO4 in the the average volume of A used.

(ii) Concentration of B in mol/dm3

(iii) Value of X. [H=1,O=16,Na=23]

SOLUTIONS:

The Equation for the Reaction: H2SO4(aq) + 2NaOH –>Na2SO4 + 2H2O

TABLE OF VALUE:

Bur Rd|Rough|1st|2nd |3rd

Final|24.00|24.30|29.30|34.30

Initial|0.00|0.00 |5.00 |10.00

Vol. |24.00|24.30|24.30|24.30

NOTE: Rough,1st titre,2nd titre,3rd titre are all in (cm3)

AVERAGE VOLUME OF ACID USED = 24.30+24.30+24.30/3 = 72.9/3 =24.30.

(Bi) A contains 0.125mol/dm3 of H2SO4(given) ie 0.125 in 1000cm3 of H2SO4 .:. X in 24.30; X = 24.30/1000 * 0.125 = 0.00303 = 0.003 mol per 24.30cm3.

(ii) Conc of A * Vol of A/Conc of B * Vol of B =nA/nB ie CAVA/CBVB = nA/nB = 0.125*24.30/CB*25 = 1/2 CB = 6.075/25 = 0.243 = 0.24mol/dm3.

(iii) B contains Xgdm3 of NaOH, Molar Mass of NaOH = 40g. Mass Conc = Mole Concentration * Molar Mass = 0.243 * 40 = 9.72g

SALT ANALYSIS:

2. (a) TEST: Xn + Distilled H2O

OBSERVATION:

White ppt, colourless solution observed.

INFERENCE:

Soluble Salt

(i) TEST: Filterate + BaCl2 + Dil HCl in excess.

OBSERVATION:

White ppt formed. Precipitate Insoluble.

INFERENCE:

SO4^2-,SO3^2-, CO3^2- Present.

TEST:

Residue + dilute HCl

OBSERVATION:

Effervescence gas evolved.

An odourless gas. Deep blue ppt.

INFERENCE:

Cu^2+ Present.

TEST:

Filterate + HNO3 + AgNO3 + NH3

OBSERVATION:

No visible reaction.

White ppt formed. Ppt dissolved in excess NH3(aq)

INFERENCE:

Cl^- present.

Cl^- confirmed.

(bi) TEST:

Filterate + BaCl2 + dil HCl

OBSERVATION:

White ppt remained insoluble.

INFERENCE:

CO3^2-, SO4^2- Present.

(ii) TEST:

A portion of (bi) + NaOH solution dropwise and in excess.

OBSERVATION:

A light blue ppt insoluble in excess of NaOH Solution.

INFERENCE: Cu^2+ present.

(NOTE THAT THE ABOVE SOLUTION WILL BE IN A TABULAR FOR i.e TEST|INFERENCE|OBSERVATION)

A mixture of NH4Cl+CuCO3 (samplC) now being called unknown sample C

Test|observation|interferent

1. Mix sampl C with water = partialy soluble = sample C is a mixture

2. A portn of the filtrate + NH3 = white ppt. = Cl- present

3. A portn of d filtrate + AgNO3 + NH3 then in excess = ppt disolvedin excess NH3 solutn = Cl- ion confirmed

4. Another portn of d filtrate + NaOH solutn then in excess + heat = a colourless gas evolved wit xteristics choking smell which turned litmus paper blue and form a dens white fume wit conc.HCl stopper = NH3 gas from NH4+

5. Residue + dil HCl = efferviscent occured, colourless gas was given off after d residue disolved which turned lime water milky = CO2 gas from CO32- or if the residue is being added witHCl = it will give clear solution the clear solution + NaOH in drop and den in excess = a pale blue ppt was form which was insoluble in excess NaOH solutn =Cu2+ is present

7. The mixture above (pale blue ppt) + NH3 solutn in drop and den in excess = pale blue ppt disolved to give a deep blue solutn = Cu2+ is confirmed.