Question

# Consider a buffer solution made of 0.255 M acetic acid, CH3COOH, and 0.165 M sodium acetate,...

Consider a buffer solution made of 0.255 M acetic acid, CH3COOH, and 0.165 M sodium acetate, CH3COONa. Ka(CH3COOH) = 1.8×10-5. After addition of 0.040 moles of NaOH to 1.0 L of this buffer, the pH becomes [X].  Fill in the blank. Show the number only. Report with 2 digits after the decimal point.

0.255 M CH3COOH in 1.0 L

So, moles of CH3COOH = 0.255 M x 1.0 L = 0.255 moles

0.165 M CH3COONa in 1.0 L

So, moles of CH3COONa = 0.165 M x 1.0 L = 0.165 moles

When 0.040 moles of NaOH is added, it will react with 0.040 moles of CH3COOH to produce 0.040 moles of CH3COONa. So, Moles of CH3COOH will decrease and moles of CH3COONa will increase.

Moles after addition of NaOH are

moles of CH3COOH = 0.255 moles - 0.040 moles = 0.215 moles
moles of CH3COONa = 0.165 moles + 0.040 moles = 0.205 moles

Total volume = 1.0 L

So,

[CH3COOH] = 0.215 moles / 1.0 L = 0.215 M
[CH3COONa] = 0.205 moles / 1.0 L = 0.205 M

Now, using Henderson-Hesselbalach equation

pH = pKa + log { [salt] / [acid] }

= pKa + log { [CH3COONa] / [CH3COOH] }

Ka of CH3COOH = 1.8 x 10-5

So, pKa = - log Ka = - log(1.8 x 10-5) = 4.74

So,

pH = 4.74 + log (0.205 / 0.215)

= 4.74 + log (0.953)

= 4.74 - 0.02

= 4.72

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