Question

In: Chemistry

A. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal...

A. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of

0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient.


Ka of acetic acid = 1.8 x 10^-5

Acetic acid/sodium acetate buffer

Acetic acid .1M

Sodium acetate .1M

Solutions

Expert Solution

first consider HCl added to pure water

we know that

moles = molarity x volume (L)

so

moles of HCl added = 0.1 x 0.25 x 10-3 = 2.5 x 10-5

now

final volume = 10 + 0.25 = 10.25 ml

now

[HCl ] = moles / volume (L)

[HCl] = 2.5 x 10-5 / 10.25 x 10-3

[HCl ] = 2.44 x 10-3

we know that

HCl is a very strong acid

so

100 % dissociation

HCl --> H+ + Cl-

so

[H+] = [HCl] dissocaited = 2.44 x 10-3

now

the hydronium ion concentration is 2.44 x 10-3


2)

now consider the buffer

the buffer contains an acid , CH3COOh and base Ch3COO-

so

when HCl is added

it reacts with the base CH3C00-

so

the reaction is

H+ + CH3C00- --> CH3COOH

now

initially

moles of CH3C00- = 0.1 x 10 x 10-3 = 1 x 10-3

moles of Ch3COOH = 0.1 x 10 x 10-3 = 1 x 10-3

now

moles of HCl added = 0.25 x 0.1 x 10-3 = 0.025 x 10-3

consider the reaction

H+ + Ch3OO- --> CH3COOH

moles of CH3C00- reacted = moles of H+ added = 0.025 x 10-3

moles ofo CH3C00- remaining = 0.975 x 10-3

moles of CH3COOH formed = moles of H+ added = 0.025 x 10-3

moles of CH3COOH finally = 1.025 x 10-3


now

pH = pKa + log [ CH3C00- / Ch3COOH]

pH = -log 1.8 x 10-5 + log [ 0.975 x 10-3 / 1.025 x 10-3 ]

pH = 4.723

now

-log [H+] = 4.723

[H+] = 1.89 x 10-5


Related Solutions

A buffer contains 0.150M acetic acid and 0.105M in sodium acetate. a) Calculate the pH of...
A buffer contains 0.150M acetic acid and 0.105M in sodium acetate. a) Calculate the pH of the buffer. b) What is the volume of 6.0M NaOH must be added to raise the pH of 100.0 mL of the buffer by 1.5 pH units? c) Calculate the pH when 2.0 mL of 6.0M is added to 100.0 mL to the buffer. please show work. question is due before 11PM
A beaker with 115 mL of an acetic acid buffer with a pH of 5.000 is...
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 0.470 MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. please show all work!
A beaker with 130 mL of an acetic acid buffer with a pH of 5.00 is...
A beaker with 130 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.20 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. Express answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.
A beaker with 145 mL of an acetic acid buffer with a pH of 5.000 is...
A beaker with 145 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 mol L−1. A student adds 4.90 mL of a 0.340 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
A beaker with 135 mL of an acetic acid buffer with a pH of 5.000 is...
A beaker with 135 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 5.30 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 180 mL of an acetic acid buffer with a pH of 5.00 is...
A beaker with 180 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M . A student adds 6.20 mL of a 0.440 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is...
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 mol L−1. A student adds 6.60 mL of a 0.300 mol L−1 HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.
A beaker with 105 mL of an acetic acid buffer with a pH of 5.000 is...
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 8.90 mL of a 0.350 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...
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is...
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.
A beaker with 105 mL of an acetic acid buffer with a pH of 5.000 is...
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 8.70 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.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT