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You need to prepare an acetate buffer of pH 5.36 from a 0.900 M acetic acid solution and a 2.62 M KOH solution. If you have 680 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.93? The pKa of acetic acid is 4.76.

Respuesta :

Answer:

Explanation:

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

5.36 = 4.86 + log [ CH₃COO⁻ ] / [CH₃COOH ]

log [ CH₃COO⁻ ] / [CH₃COOH ]  = .5

[ CH₃COO⁻ ] / [CH₃COOH ]  = 3.16

moles of CH₃COOH = .680 x .9 = .612 M

Let  x mole of KOH is required

x /( .612 - x ) = 3.16

x = 1.933 - 3.16 x

x = .46488

.46488 moles of KOH will be required

volume required be V

v x 2.62 =  .46488

v = .1774 L

= 177.4 mL

177.4 mL  of  2.62 M KOH will be required .

The number of milliliters of the KOH solution is 177.4 mL

Calculation of the number of milliliters of the KOH solution:

Since we know that

H = pKa + log [ CH₃COO⁻ ] / [CH₃COOH ]

5.36 = 4.86 + log [ CH₃COO⁻ ] / [CH₃COOH ]

Now

log [ CH₃COO⁻ ] / [CH₃COOH ]  = .5

[ CH₃COO⁻ ] / [CH₃COOH ]  = 3.16

Therefore,

moles of CH₃COOH = .680 x .9 = .612 M

here we assume  x mole of KOH

So,

x /( .612 - x ) = 3.16

x = 1.933 - 3.16 x

x = .46488

Now the volume required is

v x 2.62 =  .46488

v = .1774 L

= 177.4 mL

Learn more about solution here: https://brainly.com/question/24867768

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