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]
The Equation for the Reaction: H2SO4(aq) + 2NaOH –>Na2SO4 + 2H2O
TABLE OF VALUE:
Bur Rd|Rough|1st|2nd |3rd
Initial|0.00|0.00 |5.00 |10.00
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
2. (a) TEST: Xn + Distilled H2O
White ppt, colourless solution observed.
(i) TEST: Filterate + BaCl2 + Dil HCl in excess.
White ppt formed. Precipitate Insoluble.
SO4^2-,SO3^2-, CO3^2- Present.
Residue + dilute HCl
Effervescence gas evolved.
An odourless gas. Deep blue ppt.
Filterate + HNO3 + AgNO3 + NH3
No visible reaction.
White ppt formed. Ppt dissolved in excess NH3(aq)
Filterate + BaCl2 + dil HCl
White ppt remained insoluble.
CO3^2-, SO4^2- Present.
A portion of (bi) + NaOH solution dropwise and in excess.
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
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.