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Ugo [173]
4 years ago
12

An analytical chemist is titrating of a solution of propionic acid with a solution of . The of propionic acid is . Calculate the

pH of the acid solution after the chemist has added of the solution to it.
Chemistry
1 answer:
motikmotik4 years ago
3 0

Answer:

Heres my attempt at this and hope it helps friend

.1372L * (.83M/L) = .114 mols propanoic acid

.06862L * (1.1M/L) = 0.0755 mols NaOH

using the henderson hasselbach equation

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

so

ph = 4.89 + log(.0755/.114) = 4.72

Explanation:

Identify at least three reasons the Articles of Confederations failed as a governing document. In your opinion, evaluate which defect was most debilitating, using evidence and your knowledge of American government to justify your position.

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Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2
quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

<u>The formula used for isothermally irreversible expansion is :</u>

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done

p_{ext} = external pressure

V_1 = initial volume of gas

V_2 = final volume of gas

<u>The expression used for work done in reversible isothermal expansion will be,</u>

w=-nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 25^oC=273+25=298K

V_1 = initial volume of gas

V_2 = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(2.5atm)\times (10-1)L

w=-22.5L.atm=-22.5\times 101.3J=-2279.25J

(2) A free isothermal expansion from 1 L to 100 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})

w=-11409.6J

(3) A reversible isothermal expansion from 0.5 L to 4 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})

w=-5151.97J

(4) A reversible isothermal expansion from 0.5 L to 40 L.

w=-nRT\ln (\frac{V_2}{V_1})

w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})

w=-10856.8J

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

w=-p_{ext}(V_2-V_1)

w=-(0.5atm)\times (100-1)L

w=-49.5L.atm=-49.5\times 101.3J=-5014.35J

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0
3 years ago
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