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Each of the observations listed was made on a different solution. Given that obs

ID: 921831 • Letter: E

Question

Each of the observations listed was made on a different solution. Given that observations, state which ion studied in this experiment is present. If the lest is not definitive, indicate that with a question mark. Addition of 6 MNaOH and Al to the solution produces a vapor that turns red litmus blue. Addition of 6 MHCl produces an effervescence. Addition of 6 MHNO_3 plus 0.1 MAgNO_3 produces a precipitate. Addition of 6 MHCl plus 1 MBaCl_2 produces a precipitate. Addition of 6 MHNO_3 plus 0.5 M(NH_4)_2 MoO_4 produces a precipitate. An unknown containing one or more of the Ions studied in this experiment has the following properties: no effect on addition of 6M acetic acid MFe(NO_3)_3; no effect on addition of 6 M HNO_3 and 0.1 MAgNO_3; white precipitate on addition of 6 M HCl and 1M BaCl_2; yellow precipitate on addition of 6 M HNO_3 and (NH_4)_2MoO_4; On the basis of this information which ions are present and which are absent

Explanation / Answer

1(a): When 6M NaOH reacts with Al, it forms Al(OH)4- which is basic in nature and hence turns red litmus to blue. Hence the ion present is OH-.

(b): Efervescence is produced when gases like CO2 is liberated during the reaction. Hence addition of HCl conforms the presence of effervescence conforms the presence carbonate ion (CO32-) that libertes CO2 gas when reacts with HCl

CaCO3 + HCl --- > CaCl2 + H2O + CO2(effervescence)

Hence the ion present is carbonate ion (CO32-)

(c): AgNO3 when reacts with acetate ion (CH3COO-) it forms a white precipitate of silver acetate (CH3COOAg).

Hence the ion present is CH3COO-

(d): When HCl and BaCl2 react with sulphate ion(SO42-), a white precipitate of BaSO4 is formed. Hence the ion present is ulphate ion(SO42-)

Ba2+(aq) + SO42-(aq) ---- > BaSO4(s)

(e): When HNO3, H3PO4 and (NH4)2MoO4 react they yellow colored precipitation of ammonium phosphomolybdate ((NH4)3PO4•12MoO3). Hence the ion present is PO43- ion

H3PO4 + (NH4)2MoO4 + HNO3 ---- > (NH4)3PO4•12MoO3 (yellow ppt) + NH4NO3 + H2O

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