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A solution containing 0.400 g of a diprotic acid requires 40.00cm^3 of 0.100 mol dm^3...
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A diprotic acid is a acid that have the capability to provide two `H^+ ` ions to the solution such as `H_2SO_4` . Let us take the acid as `H_2A`.
`2NaOH+H_2A+ rarr Na_2A+2H_2O`
Amount of NaOH used for reaction `= 0.1/1000xx40 = 0.004mol`
Since 0.004moles of NaOH reacted the amount of diprotic acid moles reacted will be half of NaOH reacted.
`NaOH:H^+ = 2:1`
Amount of diprotic acid reacted `= 0.004/2 = 0.002mol`
Relative molecular mass
`= 0.4/0.002 `
`= 200 `
So the answer is (3)
Posted by jeew-m on May 7, 2013 at 6:59 AM (Answer #1)
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