The following observations, A-D, are made about reactions of sodium hydroxide, N

Question

The following observations, A-D, are made about reactions of sodium hydroxide, NaOH. Discuss the chemical processes involved in each case. Use principles from acid-base theory, oxidation-reduction and bonding and/or intermodular forces to support your answers. When a few drops of 3 M NaOH solution are added to 10 mL of 0.50 M aluminum chloride, AICI3, a white precipitate forms. When excess NaOH solution is added to the mixture containing the precipitate, the precipitate dissolves. When pellets of NaOH are added to water, there is substantial increase in the temperature of the system as the pellets dissolve. When 15 mL of 1 M sodium hydroxide, NaOH, is added to 10 mL of 1 M hydrochloric acid, HC1, the resulting mixture has pH greater than 7. When 15 mL of 1 M NaOH is added to 10 mL of 1 M phosphoric acid, H3PO4, the resulting mixture has pH less than 7. When a solution of 0.10 M NaOH is exposed to the atmosphere for several days, both its volume and its pH decrease.

Explanation / Answer

A) when NaOH is added to AlCl3 solution Al(OH)3 precipitate is formed . and then again excess NaOH added this precipitate dissolves and forms sodium meta aluminate

3NaOH + AlCl3 --------------------> Al(OH)3 (s) + 3 NaCl

2Al(OH)3 + 2NaOH -------------------------> 2NaAlO2 + 4 H2O

B) NaOH pellet are added to water exothermic reaction takes place heat is released.

NaOH(s) + H2O(l) => Na+ + OH- + H20 + HEAT

in this NaOH is ions bonds broken NaOH-----------------> Na+ + OH-

H2O hydeogen bond broken   H2O -------------------> H+ + OH-

these two are endothermic processs. but new electrostatic attraction between solvent - solute (NaOH) are developed. it is exothermic reaction here realesd energy is dominent , so overall due to ion -dipole attraction it is exothermic

C) 15ml, 1 M NaOH + 10 ml ,1M HCl

millimoles of NaOH = 15x1= 15

millimoles of HCl = 10x1= 10

base is dominent

[OH-] = (15-10)/total volume

           = 5/(10+15)

           = 0.2

pOH= -log[OH-] = -log(0.2) = 0.698

pH + pOH =14

pH = 13.3

pH>7

H3PO4 millimoles = 10 x1 = 10

acid millimoles = 3 x 10 =30

here acid > base

[H+] = (30-15)/(25) = 0.6

pH = 0.22

pH < 7

D) when NaOH exposed to atmosphere due its hygroscopic nature is converted into liquid and then starts to evaporate. so volume decrease. no of moles OH- decrease in NaOH decrease. so pOH increases and pH decrease

  

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