Question

A beaker with 1.50×10² mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M . A student adds 7.00 mL of a 0.390 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

let acetic acid be represented as HA, conjugate base as A-

given HA+A- = 0.1

moles of HA and A tohether [ HA] +[A-]= Molarity* volume (L)= 0.1*1.5*100/1000=0.1*0.15= 0.015

given pH of the buffer=5

pH= pKa+ log [A-]/[HA]

5= 4.74+log [A-]/[HA]

[A-]/[HA] =1.82. [A-]= 1.82HA

but [A-]+ [HA]=0.1

1.82[HA]+ [HA]=0.1, [HA]=0.1/2.82= 0.035 M, [A-]= 0.1-0.035=0.065

moles of HA in 1.5*100ml of buffer= 0.035*150/1000 =0.005, moles of A- =0.065*150/1000=0.00975

when 7ml of 0.39MHCl is added, moles of HCl addded= 0.39*7/1000 =0.00273

this will reacts with sodium acetate to give more moles of HA

CH3COONa+HCl------->CH3COOH+ NaCl, molar ratio of CH3COONa: HCl = 1:1, actual ratio =0.00975:0.00273. this suggest exccess reactant is sodium acetate and all the HCl reacts. Moles of CH3COOH formed= 0.00273

moles of sodium acetate consumed= 0.00273

moles of acetic acid = moles of aceitc acid initially present + moles of acetic acid produced due to reaction= 0.00273+0.005=0.00773, moles of sodium acetate remaining = 0.00975-0.00273= 0.00702

volume after mixing =1.5*100+7= 157 ml= 0.157L (1000ml=1L)

hence HA now = [AH]= 0.00773/0.157 =0.049M, [A-] = 0.00702/0.157 =0.045M

pH= pKa+log[A-]/[HA]= 4.74+log (0.045/0.049)= 4.70

change in pH= 5-4.7=0.3