0.01 M HC1 with 0.01 M NaOH: The relative conductivities of the solutions tested

Question

0.01 M HC1 with 0.01 M NaOH: The relative conductivities of the solutions tested are as follows: 0.01 M HC1:_____ 0.01 M NaOH:_____ Mixture:_______ Do the reactants exist as molecules or ions? ________ Does the conductivity change indicate the creation of more ions or a decrease in the number of ions?_________ The total ionic equation for the reaction is_________ The net ionic equation for the reaction is_________ Interpret any changes in the conductivity of the solutions, before and after mixing, in accordance with the preceding equations: 0.1 M HC_2H_3O_2 with 0.1 M NH_3: the relative conductivities of the solutions tested are as follows: 0.1 M HC_2H_3O_2:_____ 0.1 M NH_3 measured in parallel:_____ Mixture:_____

Explanation / Answer

at T = 20°C, and at a concentation of 0.01 M of HCl, expect a condutivity coefficient of 0.0147

for the NaOH, expect a similar conductivity, i.e. 0.0152 approx.

when we mix this, neutralizatin will occur so

NaOH + HCl = H2O + NaCl is formed

conductivity is now

clearly, since this is pretty high, it should be in IONIC state...

note that HCl ionizes 100% to form H+ and Cl-

note that NaOH ionizes 100% to form Na+ and OH-

then...

conductivity changes indicating that the H+ and OH- ions which were accutned before, are not any longer present in solution, so it decreases, showing only Na+ and Cl- ions are left

The total ionic equiton

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) --> H2O(l) +  Cl-(aq) + Na+(aq)

net ionic

H+(aq) + OH-(aq) --> H2O(l)

ignore Na+ and Cl- since those are spetator ions

before:

there are ions of H+ and Cl- completely dissociated, so they are high in conductivity

there are ions of Na+ and OH- completely dissociated, so they are high in conductivity

after mix, OH- and H+ form H2O so it is molecular, not ionic, there is a decrease in conductivity

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