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A student measured 20.0 mL of carbonated drink solution into a 50 mL beaker and...
A student measured 20.0 mL of carbonated drink solution into a 50 mL beaker and titrated it with sodium hydroxide solution that is 0.04M...
...The initial sodium hydroxide reading was 30.7 mL while the final reading was 52.0 mL. How many grams of the main component of carbonated drinks were present in 200 mL solution?
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Titration of NaOH and carbonated water...
`CO_2 + H_2O -> H_2CO_3`
`2 NaOH + H_2CO_3 -> Na_2CO_3 + 2 H_2O`
2 mole of NaOH react with `H_2CO_3` ....
Volume of `H_2CO_3` `(V_1)` = 20ml
Molarity of `H_2CO_3 (M_1) = ?`
moles of `H_2CO_3 (n_1) = 1`
Volume of `NaOH (V_2) = 52-30.7 = 21.3 ml`
Molarity of `NaOH (M_2) = 0.04 M`
moles of `NaOH (n_2) = 2`
Formula of dilution...
`(M_1V_1)/(n_1) = (M_2V_2)/(n_2)`
`M_1 = (M_2V_2n_1)/(n_2V_1)`
`M_1 = (0.04*21.3*1)/(2*20)`
`M_1 = 0.0213 M`
Now we got molarity of `H_2CO_3` , now we have to find the amount of `H_2CO_3` in 200ml.
Molarity = moles/volume in litres
`Molarity = ((mass)/(molar mass)) * ((1000)/(V in ml))`
`0.0213 = ((mass)/(62.03)) * ((1000)/(200))`
`Mass = (0.0213*62.03*200)/(1000)`
Mass = 0.264 g
Therefore 0.264 g of `H_2CO_3` present in 200 ml.
Posted by sanjeetmanna on August 24, 2012 at 2:44 AM (Answer #1)
We have the cocncentration of NaOH as 0.04 M
Now the volume of NaOH consumed in the reaction = final reading - initial reading = 52.0mL-30.7mL = 21.3mL
[nb:NOw in carbonates drinks, the CO2 dissolves froming H2CO3.
CO2 + H2O======> H2CO3]
the titration reaction would be
H2CO3 + 2 NaOH ======> Na2CO3 + H2O
M1 = molarity 0f NaOH = 0.04 M : V1 = volume of NaOH = 21.3 mL : n1= co-efficient of NaOH in the balanced reaction = 2
M2 = molarity 0f H2CO3 = ? : V2 = volume of H2CO3 = 20 mL : n2= co-efficient of H2CO3 in the balanced reaction = 1.
The titration formula is
M2=0.426/20 = 0.0213 M H2CO3
So the concentration of H2CO3 is 0.0213 M.
Let's find the number of moles of H2CO3 present in 200 mL of the solution/drinks
Now molarity = moles/volume (liter)
Moles = molarity * volume
Moles of H2CO3 = 0.02132 M * ( 200/1000)L
Moles of H2CO3 =0.00426 moles
Moles = mass/molar mass
Mass = moles* molar mass
the molar mass of H2CO3 is 62.04 g/mol
Mass of H2CO3 = 0.00426 mol * 62.04 g/mol = 0.264 g H2CO3
Posted by chaobas on July 13, 2012 at 9:25 PM (Answer #2)
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