You have recently joined the EPA and your immediate task is to evaluate the acid

Question

You have recently joined the EPA and your immediate task is to evaluate the acidity levels in the onandaga Lake. For this purpose you need to prepare 100 mL of a .200 M acetate buffer, pH 5.00, starting from solid NaOAC and .1 M solution of HOAc. Pleas record the number of mL of .100 M HOAc, and the number of grams NaOAc.
Values given: -Formula weight of sodium acetate= 82 g/mol -pKa of acetic acid= 4.74
Use the Henderson equation: pH=pKa + log( [salt]/[acid] )
You have recently joined the EPA and your immediate task is to evaluate the acidity levels in the onandaga Lake. For this purpose you need to prepare 100 mL of a .200 M acetate buffer, pH 5.00, starting from solid NaOAC and .1 M solution of HOAc. Pleas record the number of mL of .100 M HOAc, and the number of grams NaOAc.
Values given: -Formula weight of sodium acetate= 82 g/mol -pKa of acetic acid= 4.74
Use the Henderson equation: pH=pKa + log( [salt]/[acid] )

Values given: -Formula weight of sodium acetate= 82 g/mol -pKa of acetic acid= 4.74
Use the Henderson equation: pH=pKa + log( [salt]/[acid] )

Explanation / Answer

100 mL of a .200 M acetate buffer = 0.100 * 0.200 = 0.02 mole.

let there are x mole NaOAc and (0.02 - x) mole HOAc .

Henderson equation:

pH=pKa + log( [salt]/[acid] )

5.00 = 4.74 + log [x / (0.02 - x)]

x / (0.02 - x) = 1.82

x = 0.0129 mole.

so moles of NaOAc = 0.0129 mole.

0.0129 mole NaOAc = 0.0129 * 82 = 1.06 gm NaOAc.

moles of HOAC = (0.02 - 0.0129) = 7.1 * 10^-3 mole.

volume of HOAC = 1000 * 7.1 * 10^-3 / 0.100 = 71 ml

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