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Kazeer [188]
4 years ago
6

A eudiometer contains a 65.0 ml sample of a gas collected

Chemistry
1 answer:
SCORPION-xisa [38]4 years ago
6 0

Answer:

53.1 mL

Explanation:

Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.

For the ideal gas law:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.

At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:

P1 = 92.5 - 2.8 = 89.7 kPa

T1 = 23°C + 273 = 296 K

89.7*65/296 = 101.325*V2/273

101.325V2 = 5377.45

V2 = 53.1 mL

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Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

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Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

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