Given that Ka for HOCl is 3.5x10^-8 and the Ka for HF is 7.2x10^-4 (both at 25 C

Question

Given that Ka for HOCl is 3.5x10^-8 and the Ka for HF is 7.2x10^-4 (both at 25 C), identify the acid-base conjugate pairs and determine in which way the equilibrium will persist. HOCl (aq) + F- (aq) <---> HF (aq) + OCl- (aq)
Determine whether the following compounds produce an acidic, basic, or neutral solution when dissolved in water: Fe(NO3)3 AlCl3 Na2CO3 BaClO4 LiCN Given that Ka for HOCl is 3.5x10^-8 and the Ka for HF is 7.2x10^-4 (both at 25 C), identify the acid-base conjugate pairs and determine in which way the equilibrium will persist. HOCl (aq) + F- (aq) <---> HF (aq) + OCl- (aq)
Determine whether the following compounds produce an acidic, basic, or neutral solution when dissolved in water: Fe(NO3)3 AlCl3 Na2CO3 BaClO4 LiCN HOCl (aq) + F- (aq) <---> HF (aq) + OCl- (aq)
Determine whether the following compounds produce an acidic, basic, or neutral solution when dissolved in water: Fe(NO3)3 AlCl3 Na2CO3 BaClO4 LiCN

Explanation / Answer

the strongest acid is HF, that is, HF <-> H+ and F- is more favoured than the one of HOCl

then

HOCl (aq) + F- (aq) <---> HF (aq) + OCl- (aq) is NOT favoured

HF (aq) + OCl- (aq) <---> HOCl (aq) + F- (aq)

acid --> HF

bae = OCl-

conjugate acid = HOCl

conjugate base = F-

2.

Fe(NO3)3 --> acidic, since Fe+ 2H2O <-> F(OH)2 + 2H+

AlCl3 --> acidic, since Al+3 + 3H2O <-> Al(OH)3 + 3H+

Na2CO3 --> basic since CO3-2 + H2O <-> HCO3- + OH-

BaClO4 --> acidic since Ba+2 + H2O <-< BA(OH)2 + 2H+

LiCN --> basic since CN- + H2O <-> HCN + OH-

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