A.) Calculate the pH of a buffer that is 0.190M in NaHCO3 and 0.315M in Na2CO3.

Question

A.) Calculate the pH of a buffer that is 0.190M in NaHCO3 and 0.315M in Na2CO3.

B.) Calculate the pH of a solution formed by mixing 65 mL of 0.35M NaHCO3 with 75 mL of 0.27M Na2CO3.

C.)You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa).

What is the pH of the benzoic acid solution prior to adding sodium benzoate?

ow many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Explanation / Answer

A) we know that for a buffer

pH = pKa + log [conjugate base / acid ]

so

pH = pKa + log [Na2C03 / NaHC03]

pKa for HC03- is 10.32

so

using given values

we get

pH = 10.32 + log [0.315 / 0.19]

pH = 10.54

so

the pH of the buffer is 10.54

B)

we know that

moles= molarity x volume (L)

so

moles of NaHC03 = 0.35 x 65 x 10-3 = 22.75 x 10-3

moles of Na2C03 = 0.27 x 75 x 10-3 = 20.25 x 10-3

final volume = 65 + 75 = 140 ml

now

final conc of NaHC03 = moles / volume (L)

final conc of NaHC03 = 22.75 x 10-3 / 140 x 10-3 =0.1625 M

final conc of Na2C03 = 20.25 x 10-3 / 140 x 10-3 = 0.144643 M

now

pH = pKa + log [Na2C03 / NaHC03]

pKa for HC03- is 10.32

so

using given values

we get

pH = 10.32 + log [0.144643 / 0.1625]

pH = 10.27

so

the pH of the buffer is 10.27

C)

before adding sodium benzoate :

Ka of benzoic acid is 6.28 x 10-5

it is a weak acid

for weak acids

[H+] = sqrt ( Ka x C)

so

[H+] = sqrt ( 6.28 x 10-5 x 0.02)

[H+] = 1.12 x 10-3

pH = -log [H+]

pH = -log 1.12 x 10-3

pH = 2.95

so the pH prior to addign sodium benzoate is 2.95

we know that

for buffers

pH = pKa + log [conjugate base / acid ]

pH = -log Ka + log [C6H5COONa / C6H5COOH]

so

4 = -log 6.28 x 10-5 + log [C6H5COONa / C6H5COOH]

log [C6H5COONa / C6H5COOH] = 0.202

[C6H5COONa / C6H5COOH] = 0.628

now

we know that

moles = conc x volume

as the final volume is same for both , volumes cancel out

ratio of conc = ratio of moles

so

moles C6H5COONa / moles of C6H5COOH = 0.628

moles of C6H5COONa = 0.628 x moles of C6H5COOH

now

we know that

moles = conc x volume

so

moles of C6H5COOH = 0.02 x 1.5 = 0.03

so

moles of C6H5COONa = 0.628 x 0.03 = 0.01884

now

mass = moles x molar mass

molar mass of C6H5COONa is 144 g/mol

so

mass of C6H5COONa = 0.01884 x 144

mass of C6H5COONa = 2.713

so

2.713 grams of C6H5COONa is required

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