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mihalych1998 [28]
4 years ago
12

Suppose 50 mL of KOH with unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4

is dripped into the KOH solution. After exactly 33.4 mL of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH?
Chemistry
1 answer:
harkovskaia [24]4 years ago
6 0

Answer:

0.13 M

Explanation:

Since 1 mole of H2SO4 reacts with 2 moles of KOH, there are twice as many moles of KOH at the equivalence point as H2SO4

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Answer:

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Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g)         Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g)                        Reaction 2

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1 mol CH₃CH₂OH ______  46.0684 g

x                            ______   50.0 g

x = 1.085 mol  CH₃CH₂OH

1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

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100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

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