Determination of acid or basic solution

[Sophie522]

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When sodium acetate, NaC2H3O2 is dissolved in water, the resukting solution is

A. acidic
B. basic
C. neutral
D. colligative
E. not enough information given.

Correct Answer: B

I thought the solution would be acidic because the acetate ion would acquire a proton forming acetic acid. How would you justify answer B? Thanks.

 

[CANgnome]

dissolves into sodium and ethanoic acid.

Sodium is a stable ion, so it will not do anything. By this, I mean it will NEVER go get a OH- group since NaOH is a strong base and dissociates completely.

ethanoic acid, however, is a weak acid. so initially it will be [Eth-O-] without hydrogen. IT IS A WEAK ACID, so it will go to equilibrium with its conjugate base. How do you make conjugate base? take a H from the solution! So, it lowers the [H+] in the solution making a little bit of [Eth-O-H] from [Eth-O-]. Lowering the H+ means upping the pH therefore, making the solution more basic.

 

[LuckMC11]

this is the equation u end up with:

CH3COONa + H2O ---> CH3COOH + NaOH

as u can see, NaOH is a much more stronger base than the level that CH3COOH is acidic, so it becomes basic.

hope this helps!

 

[Sophie522]

cangnome, your explanation is a bit hard for me to understand, but i think i get the gist of what you're saying.

so i can say that because ethanoic acid/ acetic acid is a weak acid, its conjugate base is strong, therefore, soln is basic?

thanks for your responses!

 

[CANgnome]

I think i worded it a bit hard to understand :X

I will try to explain again:

so from NaCH3COO dissociating we get:

Na+ and CH3COO-

now, Na+ is perfectly fine by itself since its conjugate base is a strong Base.

NaOH ---------> Na+ and OH- (nonreversible!)

there is no way for it to go back since NaOH dissociated completely in solution.



In comparison, CH3COO- is a weak base of a weak conjugate acid, so it exists in equilibrium when in a solution with its conjugate acid:

CH3COO- + H+ <----------> CH3COOH (reversible!)

So, if you dump in CH3COO- into the solution which has no CH3COOH to begin with, it will want to go to equilibrium by forming CH3COOH. How can it form CH3COOH? Grab some H+ from the solution! So it will lower [H+] and raise pH.

 

[siffe]

I hope this reaction helps you to understand what CANgnome means:

CH3COO-(aq) + H2O(l) --> CH3COOH(aq) + OH-(aq)

CH3COO- pulls one H+ from H2O, and leaves OH- left on solution.

The key is really to recognize which acids are weak acids and how they work.