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A student measured 20.0 mL of carbonated drink solution into a 50 mL beaker and...

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spock14 | Student, Undergraduate | eNoter

Posted July 13, 2012 at 4:14 PM via web

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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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sanjeetmanna | College Teacher | (Level 3) Assistant Educator

Posted August 24, 2012 at 2:44 AM (Answer #1)

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Titration of NaOH and carbonated water...

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` ....

Given

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.

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chaobas | College Teacher | Valedictorian

Posted July 13, 2012 at 9:25 PM (Answer #2)

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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.

that is 

CO2 + H2O======> H2CO3]

Now

the titration reaction would be

H2CO3  + 2 NaOH ======> Na2CO3 + H2O

Let,

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 

 

(M1V1)/n1= (M2V2)/n2

(0.04*21.3)/2=(M2*20)/1

0.426=M2*20

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 * volum

Moles of H2CO3 = 0.02132 M * ( 200/1000)L

Moles of H2CO3 =0.00426 moles

 

we have 

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

 

 

 

 

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