Unifying Concepts Question Help Watch

grapesugar
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Report Thread starter 12 years ago
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hey..so i was doing the jan 06 paper and im stuck...

Hydroiodic acid, HI is a strong acid. this is how it is prepared.

A mixture of 480g of iodine and 600cm3 of water was put into a flask. the mixture was stirred and hydrogen sulphide gas, H2S(g) was bubbled through for several hours.

the mixutre became yellow as sulphur seperated out. the sulphur was filtered off and the solution was purified by fractional distilation. A fraction of HI (aq) was collected containing 440g of HI ina total volume of 750cm3.

please note: when i write H2S the two is a small one (subscribt) i dont know how to make it come up here..

anyway here are the questions:

(i) determine the percentage yield of hydroiodic acid (3marks)
(ii) calculate the pH of the hydroiodic acid fraction (2marks)

PLEASE help meeeeeeeeeeee
:suith:
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charco
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H2S + I2 --> 2HI + S

so calculate the moles of iodine used = 480/254 = 1.89

in an ideal world this would make double the number of moles of HI = 3.78 = 483.4g

but only 440g was made so this represents a % yield of 440/483.4 x 100 = 91%

in 750 ml of water there are 3.78 x 91/100 moles = 4.59 moles per litre

HI is a strong acid 100% dissociated = 4.59 moles per litre H+ ions

pH = -log10[H+]

pH = -0.66
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grapesugar
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wow..thankyou so much!!!!!!!!
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