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10.0 cm^3 of 1.00 mol dm^3 HCl neutralized 20.0 cm^3 of a NaOH solution. What is the...

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kezze | Student, Grade 11 | Valedictorian

Posted March 15, 2013 at 12:43 PM via web

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10.0 cm^3 of 1.00 mol dm^3 HCl neutralized 20.0 cm^3 of a NaOH solution. What is the molarity of the NaOH?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted March 15, 2013 at 2:58 PM (Answer #1)

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This problem is an acid-base neutralization reaction meaning that the moles of the acid are equivalent to the moles of the base. 

The reaction equation for this problem can be written as:

`HCl + NaOH -> NaCl + H_2O`

Remember that

`Molarity (M) = (mol es of Solute)/(Liters of solution)`  

which can also be written as:

`Molarity (M) = (mol es of Solute)/(dm^3 of solution)`

since

`1 Liter = 1 dm^3`

 

Given:

`M_(NaOH)` = ?

`dm^3` `NaOH =` `20.0 cm^3 (1 dm^3)/(1000 cm^3) = 0.0200 dm^3`

`M_(HCl)` = `1.00 (mol)/(dm^3)`

`dm^3` `HCl =` `10.0 cm^3 (1dm^3)/(1000cm^3) = 0.0100dm^3`

Solution:

`mol es NaOH = mol es HCl`

`(M_(NaOH))*(dm^3 _N_a_O_H) = (M_(HCl))*(dm^3 _H_C_l)`

`(M_(NaOH))*(0.0200) = (1.0)*(0.0100)`

`M_(NaOH) = ((1.0)*(0.0100))/(0.0200)`

`M_(NaOH) = 0.500 (mol)/(dm^3)`

Th molarity of the NaOH is 0.500mol/dm^3.

 

 

Sources:

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