Chemistry, 10.07.2019 01:00 kadenh420

What are the concentrations of acetic acid (pka = 4.76) and acetate in a buffer solution of 0.10 m at ph 4.9?

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Answer from: shunna33

Solution:

We have to use the Henderson-Hasselbalch equation: for this calculation

Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity by using pKa, the negative log of the acid dissociation constant in biological and chemical systems. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid-base reaction.

The equation is given by:

Here, [HA] is the molar concentration of the un dissociated weak acid, [A⁻] is the molar concentration (molarity, M) of this acid's conjugate base and pKa is −log10 Ka where Ka is the acid dissociation constant, that is:

pH = pKa + log([A^-]/[HA])

We look up the pKa for acetic acid:

pKa = 4.76

Let x = molarity of AcO^- and y = molarity of AcOH: Then we have the following two equations in two unknowns:

(1) x + y = 0.10 M

and

(2) 4.9 = 4.76 + log(x/y)

Further calcite the value of x and y by algebraic method and get the answer.

Answer from: raemcbride5678

Solution:

We have to use the Henderson-Hasselbalch equation: for this calculation

The equation is given by:

pH = pKa + log([A^-]/[HA])

We look up the pKa for acetic acid:

pKa = 4.76

Let x = molarity of AcO^- and y = molarity of AcOH: Then we have the following two equations in two unknowns:

(1) x + y = 0.10 M

and

(2) 4.9 = 4.76 + log(x/y)

Further calcite the value of x and y by algebraic method and get the answer.

Answer from: Quest

first one is the correct answer

Answer from: Quest

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

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What are the concentrations of acetic acid (pka = 4.76) and acetate in a buffer solution of 0.10 m a...

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