Learning Goal:
To understand how buffers use reserves of conjugate acid and conjugate base to counteract the effects of acid or base addition on pH.
A buffer is a mixture of a conjugate acid-base pair. In other words, it is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, an acetic acid buffer consists of acetic acid, CH3COOH, and its conjugate base, the acetate ion CH3COO−. Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (e.g., sodium acetate NaCH3COO).
Buffers work because the conjugate acid-base pair work together to neutralize the addition of H+ or OH− ions. Thus, for example, if H+ ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of H+with the conjugate base:
H++CH3COO−→CH3COOH
Similarly, any added OH− ions will be neutralized by a reaction with the conjugate acid:
OH−+CH3COOH→CH3COO−+H2O
This buffer system is described by the Henderson-Hasselbalch equation
pH=pKa+log[conjugate base][conjugate acid]
A beaker with 1.10×102 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 7.50 mL of a 0.490 M HCl 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.
Sol.
As Total molarity of buffer = 0.1 M
So , [CH3COOH] + [CH3COO-] = 0.1 ( equation 1 )
Initial pH = 5
Also , pKa = 4.74
So , Using Henderson - Hasselbalch equation ,
initial pH = pKa + log ( [CH3COO-] / [CH3COOH] )
5 = 4.74 + log( [CH3COO-] / [CH3COOH] )
log( [CH3COO-] / [CH3COOH] ) = 5 - 4.74 = 0.26
[CH3COO-] / [CH3COOH] = 100.26 = 1.8197
[CH3COO-] = 1.8197 [CH3COOH]
Putting this in equation number 1 ,
1.8197 [CH3COOH] + [CH3COOH] = 0.1
[CH3COOH] = 0.0355 M
[CH3COO-] = 1.8197 × 0.0355 = 0.0646 M
As the volume of buffer = 1.10 × 102 mL
So , initial millimoles of CH3COOH
= 0.0355 × 1.10 × 102 = 3.905 mmol
and initial millimoles of CH3COO-
= 0.0646 × 1.10 × 102 = 7.106 mmol
Now , millimoles of HCl added = 0.490 × 7.50 = 3.675 mmol
So , after addition of HCl , millimoles of CH3COOH increases and millimoles of CH3COO- decreases
New millimoles of CH3COOH = 3.905 + 3.675 = 7.58 mmol
New millimoles of CH3COO- = 7.106 - 3.675 = 3.431 mmol
So , Using Henderson - Hasselbalch equation ,
New pH = pKa + log ( New millimoles of CH3COO- / New millimoles of CH3COOH )
= 4.74 + log ( 3.431 / 7.58 )
= 4.40
Therefore , change in pH
= New pH - Initial pH
= 4.40 - 5.00
= - 0.60
Learning Goal: To understand how buffers use reserves of conjugate acid and conjugate base to counteract...
To understand how buffers use reserves of conjugate acid and conjugate base to counteract the effects of acid or base addition on pH. A buffer is a mixture of a conjugate acid-base pair. In other words, it is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, an acetic acid buffer consists of acetic acid, CH3COOH, and its conjugate base, the acetate ion CH3COO−. Because ions cannot simply...
A beaker with 2.00×102 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 5.40 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A beaker with 1.90×102 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 8.40 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A) A beaker with 1.80×102 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 5.50 mL of a 0.340 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. B)You need to produce a buffer solution that has a pH of 5.37. You already have a solution...
Consider a buffer composed of the weak acid acetic acid (CH3COOH) and the conjugate base sodium acetate (NaCH3COO). Which pair of concentrations results in the most effective buffer (i.e. has the highest buffer capacity)? A. 0.10 M CH3COOH; 0.10 M NaCH3COO B. 0.90 M CH3COOH; 0.10 M NaCH3COO C. 0.10 M CH3COOH; 0.90 M NaCH3COO D. 0.50 M CH3COOH; 0.50 M NaCH3COO
A beaker with 105 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 5.70 mL of a 0.480 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A beaker with 185 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 7.50 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
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 8.70 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A beaker with 165 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 4.00 mL of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
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 0.330 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.