Question

1) You pipet a known amount of this standardized HCl into a 15 cm sample tube...

1) You pipet a known amount of this standardized HCl into a 15 cm sample tube and float a "boat" containing your unknown.
You hook up the sample tube and contents to the gas buret.
You shake the tube to sink the boat and to start the reaction.
When the metal unknown is completely dissolved, you read the gas buret and measure the height of the buret's meniscus above the water level.
Experimental Data:

Volume of standardized HCl used 2.99 mL
Weight of metal unknown used 47.7 mg
Gas buret reading after reaction is compete 13.17 mL
Height of meniscus above water level 11.2 cm
Atmospheric pressure 772.9 mmHg
Temperature of the water 24.9 °C

Calculate (R = 0.0821 L atm/mol K):

f) Vapor pressure of water at 24.9 °C
interpolate data at... http://genchem.rutgers.edu/vpwater.html
______________ mmHg
g) Total volume of gas released ______________ mL
h) Partial pressure of hydrogen ______________ mmHg
i) Equivalent weight of metal
based on amount of hydrogen generated
______________ g/eq

2) The remaining solution in the 15 cm sample tube containing the reaction products and the unused HCl is quantitatively transfered to a flask and titrated to its endpoint with the standard NaOH.
Experimental data:

Initial volume of buret 0.16 mL
Final volume of buret 10.12 mL

Calculate:

j) Equivalent weight based on the amount of acid used up ______________

Homework Answers

Answer #1

1).

f). Vapor pressure of water at 24.9 °C.

At 24 oC, vapor pressure = 22.377 mmHg

At 25 oC, vapor pressure = 23.756 mmHg

V.P. = 23.756 - 22.377 = 1.379 mmHg

So there is 1.379 mmHg water vapor pressure per oC.

For 0.9 rise in temperature, vapor pressure will be 9/10 times 1.379 mmHg more than vapor pressure at 24 oC.

so the vapor pressure of water at 24.9 oC = 22.377 mmHg + (9/10)*1.379

= 22.377 + 1.2411

= 23.618 mmHg

h) Partial pressure of hydrogen

Atmospheric pressure = 772.9 mmHg

Total presure = 23.618 mmHg

P = 772.9 - 23.618

= 749.282 mmHg

Height of meniscus above water level = 11.2 cm = 112 mm

Correction factor = 112 * density of water / density of Hg

= 112 * 0.9968 g/cc / 13.55g/cc

= 8.24 mmHg

Since water is not that much dense. 112 mm water doesn't put much pressure.

Now, if the bubble is 112 mm under water, the gas would be under 8.24 mm Hg more pressure. But the bubble is 112 mm above the other water level. So the gas would be under 8.24 mm Hg less pressure.

Partial pressure of hydrogen = 749.282 - 8.24

= 741.042 mmHg

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