Question

A beaker with 155 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 5.20 mL of a 0.460 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express answer numerically to two decimal places. Use a minus sign if the pH has decreased.

Answer 1

let x = concentration of acid
let y = concentration of conjugate base

x + y = 0.155 L
5.00 = 4.740+ log y/x
5.00 - 4.740 = log y/x
0.26 = log y/x
10^0.26 = y/x

y/x =1.82

1.82x = y

x + 1.82 x = 0.100
2.82 x = 0.100
x =0.0354 M ( concentration of acid )
0.100 - 0.0354 =0.0646 M (concentration conjugate base )

moles acid = 0.155 L x 0.0354 M= 0.00548 mol
moles conjugate base = 0.0636 M x 0.155 L=0.00985 mol

moles HCl = 5.20 x 10^-3 L x 0.460 M=0.002392 mol

CH3COO- + H+ ---------------->CH3COOH

AFTER ADDITION OF HCL

moles conjugate base = 0.00985 - 0.002392=0.007458 mole
moles acid = 0.002392 + 0.00548=0.00787 mole

total volume = 155+ 5.20 = 160.2 mL = 0.1602 L

concentration of acid = 0.00787 / 0.1602 =0.0491 M
concentration conjugate base = 0.007458/ 0.1602 =0.0465 M

pH = 4.740 + log 0.0465 / 0.0491=4.71 {pH = 4.740 + log con.of base / conc.of acid}

change pH = 5.00 - 4.71=0.29