101 521
Assignments Done
99.6%
Successfully Done
In September 2021

# Answer to Question #112375 in Inorganic Chemistry for Sheri

Question #112375
How many grams of sodium hydroxide (MM = 40) is dissolved in 100 cm3 of 0.25 M acetic acid if the pH is 5.0?
1
2020-04-27T01:51:54-0400

1) If we consider that Ka(CH3COOH) = 1.75*10-5, then

The initial quantity of acetic acid was

100/1000*0.25 = 0.025 moles.

Quantity of [H+] is

[H+] = 10^-pH = 10-5 = 0.00001

Considering the amount of dissociated acid as x, we rewrite:

0.00001x = 4.4*10-7 - 1.75*10-5x

2.75*10-5x = 4.4*10-7

x = 0.016 - amount of CH3COO- in solution. Concentration of H+ must be the same, because dissociation occurs according to equation:

CH3COOH <-> CH3COO- + H+

But [H+] is less than this amount. The rest amount of [H+] formed was neutralized by NaOH. So the difference between [CH3COO-] and [H+] will be amount of NaOH added.

n(NaOH) = 0.016-0.00001 = 0.01599

m(NaOH) = 0.6396 grams.

2)(Not very correct solution) Dissociation constant of CH3COOH was not given in the task. So, i guess, this task may be not about all that stuff written above, that may be too complex. So if it doesn't imply using of dissociation constants, just use this solution.

The initial quantity of acetic acid was

100/1000*0.25 = 0.025 moles.

Considering that acetic acid completely dissociates, [H+] must be 0.025 too. Now, due to adding of base, pH goes down.

pH=5, [H+] = 10^-pH = 10-5 = 0.00001.

The difference between initial H+ and [H+] will be amount of NaOH added.

n(NaOH) = 0.025-0.00001 = 0.02499

m(NaOH) = 0.9996 grams.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!