What is the molarity of HCl found in a titration where 30 ml of HCl is titrated with 50 ml of .1 M NaOH?

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Answer 1 · 2018-04-23T04:41:59

#0.17 \ "M"#

We have the neutralization reaction:

#HCl(aq)+NaOH(aq)->NaCl(aq)+H_2O(l)#

As you can see here, one mole of acid neutralizes one mole of base.

We use the concentration equation, which states that,

#c=n/v#

#n# is the number of moles

#v# is the volume of solution

Rearranging for moles, we get,

#n=c*v#

So, we have:

#n_(NaOH)=0.1 \ "M"*0.05 \ "L"#

#=0.005 \ "mol"#

Since one mole of acid neutralizes one mole of base, then we must have: #n_(HCl)=n_(NaOH)#.

And so,

#c_(HCl)=n_(HCl)/(v_(HCl))#

#=(0.005 \ "mol")/(0.03 \ "L")#

#~~0.17 \ "M"#