Question

You have 175 mL of an 0.29 M acetic acid solution. What volume (V) of 2.00 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.27? (The pKa of acetic acid is 4.76.)

Answer 1

pH = pka + log [acetate ]/[acetic acid]

5.27 = 4.76 + log [acetate ]/[acetic acid]

[acetate ] = 3.236 [acetic acid]

acetate moles = 3.236 acetic acid moles ...........(1)

we have acteic acid + OH-   ----> Acetate + H2O (l)

initial acid moles = Mx V = 0.29 x 175/1000 = 0.05075

let V be vol of NaOH , then NaOH moles = OH- moles = Mx V = 2 x V = 2V

then after reacting with NaOH acid moles = 0.05075-2V , acetate moles = 2V

we usbstitutes these in eq 1

( 2V) = 3.236 x ( 0.05075-2V)

2V = 0.164227 - 6.472V

V = 0.0194 L = 19.4 ml   , is vol of NaOH we must add to get required buffer