If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution. what is the concentration of the HCl?

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In a neutralization reaction, we can say that the number of moles of the base is equal with the number of moles of the acid. We can simply write it down as:

Moles base = moles acid

Moles NaOH = moles HCl

Looking at the given, we can solve for the moles of the base which is NaOH:

`mol es NaOH = Molarity * volume (L)`

`mol es NaOH = 0.1 (mol es)/(L) * ((54mL)/(1000mL))L` 

`mol es NaOH = (0.1) * (0.054)`

`mol es NaOH = 0.0054 mol es NaOH = mol es HCl`

` `  

Now since we already have the moles of HCl, we can solve for its concentration.

Concentration in molarity:

`Molarity = (mol es HCl)/(Liters of solution)`

`Molarity = (0.0054mol es)/((125mL)/(1000mL)L)`

`Molarity = 0.0432 (mol es)/(L)`


*we divided the volume by 1000 because the unit of the volume required for molarity is in L. (1000 mL = 1L)



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