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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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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