Answer:
The pH will change 0.16 ( from 5.00 to 4.84)
Explanation:
Step 1: Data given
volume of acetic acid buffer = 160 mL
The total molarity of acid and conjugate base in this buffer is 0.100 M
A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.
The pKa of acetic acid is 4.740
pH = 5.00
Step 2: Calculate concentration of acid
Consider x = concentration acid
Consider y = concentration conjugate base
x + y = 0.100
5.00 = 4.740 + log y/x
5.00 - 4.740 = log y/x
0.26 = log y/x
10^0.26 =1.82 = y/x
1.82 x = y
Since x+y = 0.100
x + 1.82 x = 0.100
2.82 x = 0.100
x =0.0355 M = concentration acid
Step 3: Calculate concentration of conjugate base
y = 0.100 - x
0.100 - 0.0355 =0.0645 M= concentration conjugate base
Step 4: Calculate moles of acid
Moles = volume * molarity
moles acid = 0.160 L * 0.0355 M= 0.00568 moles
Step 5: Calculate moles of conjugate base
moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles
Step 6: Calculate moles HCl
moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles
Step 7: Calculate new moles
A- + H+ = HA
moles conjugate base = 0.01032 - 0.003266 =0.007054 moles
moles acid = 0.00568 + 0.003266=0.008946 moles
Step 8: Calculate the total volume
total volume = 160 + 7.10 = 167.1 mL = 0.1671 L
Step 9: Calculate the concentration of the acid
concentration acid = 0.008946/ 0.1671 =0.0535 M
Step 10: Calculate the concentration of conjugate base
concentration conjugate base = 0.007054/ 0.1671 =0.0422 M
Step 11: Calculate the pH
pH = 4.740 + log 0.0535/ 0.0422=4.84
change pH = 5.00 - 4.84=0.16
The pH will change 0.16
