A beaker with 115 mL of an acetic acid buffer with a pH of 5.000 is sitting on a
ID: 487866 • Letter: A
Question
A beaker with 115 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 8.00 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
Part A A beaker with 115 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 8.00 mL of a 0.300 MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus sign if the pH has decreased. ADH 55 Submit Hints My Answers Give Up Review Part incorrect; one attempt remaining; Try AgainExplanation / Answer
let x= concentration acid
y=concentration congugate base
x+y=0.100
5.00= 4.760+log y/x
5.00-4.760=logy/x
0.24=logy/x
10^0.24=1.74=y/x
1.74x=y
x+1.74x=0.100
2.74x=0.100
x=0.0365M =concentration acid
y= 0.100-0.0365= 0.0635M =concentration conjugate base
moles acid =0.115L x 0.0365M
= 0.00419
Moles conjugate base = 0.0635 Mx 0.115l
=0.007302
Moles Hcl =8.00x10^-3 l x 0.300m
= 0.0024
Moles conjugate base = 0.007302 -0.0024
=0.004902
Moles acid= 0.00419+0.0024
= 0.00659
Total volume = 115+8.00 =123ml =0.123L
Concentration acid =0.00659/0.123
= 0.0535 M
Concentration conjugate base = 0.004902/0.123
= 0.0398 M
pH= 4.760+log 0.0398/0.0535
= 4.760+log 0.7439
=4.760-0.12848
= 4.63
pH =5.00-4.63
=0.37
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