Question

You have 875 mL of an 0.39 M acetic acid solution. What volume (V) of 1.70 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.36? (The pKa of acetic acid is 4.76.)

Answer 1

Solution :-

when NaOH added to acetic acid then it forms acetate ion by reacting with acetic acid

the moles of the acetic acid decreases by the same number of moles that is NaOH is added and same number of moles of acetate are formed .

initial moles of acetic acid = molarity * volume in liter

                                               = 0.39 mol per L* 0.875 L

                                              = 0.34125 mol

so using the Henderson equation we can make the following set up

pH= pka + log [conj. base moles / acid moles]

4.36 = 4.76 + log ([x]/[0.34125 mol -x])

4.36-4.76=log ([x]/[0.34125 mol -x])

-0.4 = log ([x]/[0.34125 mol -x])

solving for the x we get x= 0.0972 mol

so now lets calculate the volume of the NaOH

volume = moles / molarity

              = 0.0972 mol / 1.70 mol per L

              = 0.0572 L

0.0572 L *1000 ml / 1 L = 57.2 ml

So volume of the NaOH needed = 57.2 ml