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?
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)