OpenStudy (anonymous):

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?

OpenStudy (zale101):

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