Answer:
The pH will change 0.39
Step-by-step explanation:
Step 1: Data given
Volume of acetic acid : 180 mL
pH = 5.00
The total molarity of acid and conjugate base in this buffer is 0.100 M.
We add 8.50 mL (= 0.0085 L) of a 0.470 M HCl
pKa of acetic acid = 4.74
Step 2: Calculate initial concentrations
x = concentration acid
y = concentration conjugate base
x + y = 0.100
pH = pKa = log ([y]/[x])
5.00 = 4.74 + log y/x
0.26 = log y/x
10^0.26 =1.82 = y/x
1.82 x = y
x + 1.82 x = 0.100
2.82 x = 0.100
x = 0.0355 = concentration of acid
y = 0.0645 = concentration of conjugate base
Step 3: Calculate moles
Moles = molarity * volume
moles acid = 0.180 L * 0.0355 M= 0.00639 moles
moles conjugate base = 0.0645 M * 0.180 L=0.01161 moles
moles HCl = 8.50 * 10^-3 L * 0.470 M=0.003995 moles
A- + H+ = HA
moles conjugate base = 0.01161- 0.003995=0.007615 moles
moles acid = 0.00639 + 0.003995=0.010385moles
Step 4: Calculate total volume
total volume = 180 + 8.50 = 188.5 mL = 0.1885 L
Step 5: Calculate concentration
concentration acid = 0.010385 moles/ 0.1885 L =0.0551 M
concentration conjugate base = 0.007615/ 0.1885 =0.0404 M
Step 6: Calculate pH
pH = 4.74 + log 0.0404/0.0551=4.61
Step 7: Calculate pH change
change pH = 5.00 - 4.61=0.39