Question

A 150 mL sample of hydrochloric acid (HCl) completely reacted with 60.0 mL of a 0.100 M NaOH solution. The equation for the reaction is given below. HCl + NaOH mc011-1.jpg NaCl + H2O What was the original concentration of the HCl solution?

Answer 1

Start with the balanced chemical equation to see what are the mole ratios.
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
From the balanced equation, the mole ratio of HCl and NaOH is 1:1 So, the moles of NaOH is the same as HCl

We, already know that 60.0 mL of a 0.100 M NaOH solution is given, now we use these information to determine the moles of NaOH.
We are given the volume (60.m mL) and concentration (0.100 M) of NaOH.

Molarity=Moles/Liters
To find the mole, we use Moles=Liters x Molarity

Convert 60.mL to liters 60.0 mL(1L/1000 mL)=0.06 L of NaOH
Moles of NaOH=(0.06 L)(0.100 M)= 0.006 moles NaOH

Now, to find the concentration (M) of HCl, we use:
Molarity=Moles/Liters
Convert 150 mL sample of hydrochloric acid (HCl) into liters which becomes 0.15 L M of HCl=(0.006 mol HCL)/ (150 mL/1000)= 0.0009 M of HCl