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

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 6.60 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.

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  1. A beaker with 120mL 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 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?

    The pKa of acetic acid is 4.76.

    Chemistry Buffer Calculations

    1 Answer

    Stefan V.

    May 8, 2016

    Δ

    pH

    =

    0.29

    Explanation:

    The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.

    Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.

    So, the Henderson-Hasselbalch equation looks like this

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    a

    a

    pH

    =

    p

    K

    a

    +

    log

    (

    [

    conjugate base

    ]

    [

    weak acid

    ]

    )

    a

    a

    −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

    In your case, you have acetic acid,

    CH

    3

    COOH

    , as the weak acid and the acetate anion,

    CH

    3

    COO

    , as its conjugate base. The

    p

    K

    a

    of the acid is said to be equal to

    4.76

    , which means that you have

    pH

    =

    4.76

    +

    log

    (

    [

    CH

    3

    COO

    ]

    [

    CH

    3

    COOH

    ]

    )

    The pH is equal to

    5

    , and so

    5.00

    =

    4.76

    +

    log

    (

    [

    CH

    3

    COO

    ]

    [

    CH

    3

    COOH

    ]

    )

    log

    (

    [

    CH

    3

    COO

    ]

    [

    CH

    3

    COOH

    ]

    )

    =

    0.24

    This will be equivalent to

    10

    log

    (

    [

    CH

    3

    COO

    ]

    [

    CH

    3

    COOH

    ]

    )

    =

    10

    0.24

    which will give you

    [

    CH

    3

    COO

    ]

    [

    CH

    3

    COOH

    ]

    =

    1.74

    This means that your buffer contains

    1.74

    times more conjugate base than weak acid

    [

    CH

    3

    COO

    ]

    =

    1.74

    ×

    [

    CH

    3

    COOH

    ]

    Now, because both chemical species share the same volume,

    120 mL

    , this can be rewritten as

    n

    C

    H

    3

    C

    O

    O

    120

    10

    3

    L

    =

    1.74

    ×

    n

    C

    H

    3

    C

    O

    O

    H

    120

    10

    3

    L

    which is

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    a

    a

    n

    C

    H

    3

    C

    O

    O

    =

    1.74

    ×

    n

    C

    H

    3

    C

    O

    O

    H

    a

    a

    −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

    (

    1

    )

    So, the buffer contains

    1.74

    times more moles of acetate anions that of acetic acid.

    Now, the total molarity of the buffer is said to be equal to

    0.1 M

    . You thus have

    [

    CH

    3

    COOH

    ]

    +

    [

    CH

    3

    COO

    ]

    =

    0.10 M

    Once again, use the volume of the buffer to write

    n

    C

    H

    3

    C

    O

    O

    H

    120

    10

    3

    L

    +

    n

    C

    H

    3

    C

    O

    O

    120

    10

    3

    L

    =

    0.1

    moles

    L

    This will be equivalent to

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    a

    a

    n

    C

    H

    3

    C

    O

    O

    +

    n

    C

    H

    3

    C

    O

    O

    H

    =

    0.012

    a

    a

    −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

    (

    2

    )

    Use equations

    (

    1

    )

    and

    (

    2

    )

    to find how many moles of acetate ions you have in the buffer

    1.74

    n

    C

    H

    3

    C

    O

    O

    H

    +

    n

    C

    H

    3

    C

    O

    O

    H

    =

    0.012

    n

    C

    H

    3

    C

    O

    O

    H

    =

    0.012

    1.74

    +

    1

    =

    0.004380 moles CH

    3

    COOH

    This means that you have

    n

    C

    H

    3

    C

    O

    O

    =

    1.74

    0.004380 moles

    n

    C

    H

    3

    C

    O

    O

    =

    0.007621 moles CH

    3

    COO

    Now, hydrochloric acid,

    HCl

    , will react with the acetate anions to form acetic acid and chloride anions,

    Cl

    H

    Cl

    (

    a

    q

    )

    +

    CH

    3

    COO

    (

    a

    q

    )

    CH

    3

    COO

    H

    (

    a

    q

    )

    +

    Cl

    (

    a

    q

    )

    Notice that the reaction consumes hydrochloric acid and acetate ions in a

    1

    :

    1

    mole ratio, and produces acetic acid in a

    1

    :

    1

    mole ratio.

    Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    a

    a

    c

    =

    n

    solute

    V

    solution

    n

    solute

    =

    c

    V

    solution

    a

    a

    −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

    In your case, this gets you

    n

    H

    C

    l

    =

    0.300 mol

    L

    1

    volume in liters

    6.60

    10

    3

    L

    n

    H

    C

    l

    =

    0.001980 moles HCl

    The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain

    n

    H

    C

    l

    =

    0 moles

    completely consumed

    n

    C

    H

    3

    C

    O

    O

    =

    0.007621 moles

    0.001980 moles

    =

    0.005641 moles CH

    3

    COO

    n

    C

    H

    3

    C

    O

    O

    H

    =

    0.004380 moles

    +

    0.001980 moles

    =

    0.006360 moles CH

    3

    COOH

    The total volume of the solution will now be

    V

    total

    =

    120 mL

    +

    6.60 mL

    =

    126.6 mL

    The concentrations of acetic acid and acetate ions will be

    [

    CH

    3

    COOH

    ]

    =

    0.006360 moles

    126.6

    10

    3

    L

    =

    0.05024 M

    [

    CH

    3

    COO

    ]

    =

    0.005641 moles

    126.6

    10

    3

    L

    =

    0.04456 M

    Use the Henderson-Hasselbalch equation to find the new pH of the solution

    pH

    =

    4.76

    +

    log

    (

    0.04456

    M

    0.05024

    M

    )

    pH

    =

    4.71

    Therefore, the pH of the solution decreased by

    Δ

    pH

    =

    |

    4.71

    5.00

    |

    =

    ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

    a

    a

    0.29 units

    a

    a

    −−−−−−−−−−−−−

    Answer link

    Related topic

    Buffer Calculations

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  2. ΔpH = 0.296

    Explanation:

    The equilibrium of acetic acid (CH₃COOH) in water is:

    CH₃COOH ⇄ CH₃COO⁻ + H⁺

    Henderson-Hasselbalch formula to find pH in a buffer is:

    pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

    Replacing with known values:

    5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

    0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

    1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

    As total molarity of buffer is 0.100M:

    [CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

    Replacing (2) in (1):

    1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

    1.820[CH₃COOH] = 0.100M - [CH₃COOH]

    2.820[CH₃COOH] = 0.100M

    [CH₃COOH] = 0.100M / 2.820

    [CH₃COOH] = 0.035M

    Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

    5.40 mL of a 0.490 M HCl are:

    0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

    Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

    HCl reacts with CH₃COO⁻ thus:

    HCl + CH₃COO⁻ → CH₃COOH

    After reaction, moles of CH₃COO⁻ are:

    0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

    Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

    5.425x10⁻³ moles +  2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

    pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

    pH = 4.704

    As initial pH was 5.000, change in pH is:

    ΔpH = 5.000 - 4.740 = 0.296

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