If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached

25.5mL of 0.100M NaOH are needed to reach buffer capacity.

Explanation:

The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:

Moles sodium acetate / Moles Acetic acid = 10

The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:

HA + NaOH → H2O + NaA

That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.

The initial moles of each species is:

Acetic acid:

23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid

Sodium Acetate:

33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate

We can write the moles of each species when NaOH is added as:

Moles sodium acetate / Moles Acetic acid = 10

0.00623 moles + X / 0.00343 moles - X = 10

Where X are moles of NaOH added

Solving for X:

0.00623 moles + X = 0.0343 moles - 10X

11X = 0.0281

X = 0.00255 moles of NaOH are needed

In Liters:

0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =

the answer is the first option

explanation:

just googled it m8. not that hard

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