Question

What is the molarity of HCl found in a titration where 150 ml of HCl is titrated with 10 ml of 1 M NaOH?

Answer 1

`[HCl]-=0.0667*mol*L^-1`

Explanation:

We interrogate the reaction....

`HCl(aq) + NaOH(aq) rarr NaCl(s) + H_2O(l)`

And thus there is 1:1 stoichiometry....

And so we work out the molar quantity of `NaOH(aq)`.

`10*cancel(mL)xx10^-3*cancel(L*mL^-1)xx1.0*mol*cancel(L^-1)=0.010*mol`

And `[HCl]=(0.010*mol)/(0.150*L)=??*mol*L^-1`