H. Add about 0.5 mL (about 15 drops) of 1.0 M acetic acid solution in a small te

Question

H. Add about 0.5 mL (about 15 drops) of 1.0 M acetic acid solution in a small test tube. Add a drop or two of the phenolphthalein indicator. Carefully record the smell by smelling a stirring rod used to mix the solution. Add 20 drops of 1.0 M sodium acetate. Again, carefully record the smell after the reaction. Smells/Colors: Odoc Names: sodium ach acetate lonic Eqn. Net Ionic Eqn: I. Add about 0.5 mL (about 15 drops) of 1.0 M ammonia solution in a small test tube. Add a drop or two of the phenolphthalein indicator. Carefully record the smell by smelling a st rod used to mix the solution. Add 20 drops of 1.0 M HCI. Again, carefully record the smell after the reaction. Smells Colors

Explanation / Answer

NaC2H3O2 + HC2H3O2 --> NR

there is no reaction

this will be a buffer

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Now,

if stoichiometric ratios are given then

assume 1:1

C2H3O2- + HC2H3O2 --> HC2H3O2 + C2H3O2- K

Note that this is the net ionic equaiton expected

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