Question

In a similar conductometric titration, 7.00 mL of an unknown M of NaOH was titrated with 0.115 M HCL. The equivalence point volume was 12.5 mL of HCl. [NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)]

(b). Classify the reactants in terms of electrolytes.

(c). Classify the products in terms of electrolytes.

(d). As the reaction progresses, will the conductivity go up or down? (As products are made and reactants used up, will you have more or fewer ions in solution?)

Answer 1

NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)

1 mole         1 mole

HCl

molarity M1 = 0.115 M

volume V1 = 12.5ml

no of moles n1 = 1

NaOH

molarity M2 =

volume V2 = 7ml

no of moles n2 = 1

M1V1/n1     = M2V2/n2

M2 = M1V1n2/n1V2

   = 0.115*12.5*1/1*7 = 0.2053 M

molarity of NaOH is 0.2053M

b. NaOH is base which gives OH- ions in aquous solution NaOH-----> Na+ + OH-

HCl is acid which gives H+ ions in aquous soultion HCl----> H= + Cl-

c NaCl is salt it is stron electrolyte.H2o is week electroly.

d. conducivity is equal