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3 years ago

9

You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration

of 250 mm and a ph of 5. what concentrations of acetic acid and sodium acetate should you use

1 answer:

vovangra [49]3 years ago

7 0

<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM

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2 years ago

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

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29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

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$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) <em>Alice heats the gas at constant volume until its pressure is doubled</em>:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) <em>Bob further heats the gas at constant pressure until its volume is doubled</em>:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

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