A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH. How would you calculate the molarity of HCl in this solution?

Question

A 30 ml volume of HCl is titrated with 23 mL of 0.20 M NaOH.  How would you calculate the molarity of HCl in this solution?

Wyatt Atkins · Accepted Answer

Correctly! And according to the equation:

#HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(aq)# Moles of sodium hydroxide are equivalent to the moles of hydrochloric acid. I know (or can calculate) the moles of sodium hydroxide, and I know the equivalent quantity of #HCl# was dissolved in a #30xx10^-3L# volume. Volume (#L#) #xx# concentration (#mol*L^-1#) gives an answer in #mol#. Note that #1# #mL# #=# #1xx10^-3*L# So #(23xx10^-3cancelLxx0.20*mol*cancel(L^-1))/(30xx10^-3L)# #=# #???mol*L^-1#

Jeremiah Adams · Answer

undefined To calculate the molarity of HCl in the solution, you can use the equation for neutralization reactions:

M1V1 = M2V2

Where:
M1 = Molarity of HCl
V1 = Volume of HCl (in liters)
M2 = Molarity of NaOH
V2 = Volume of NaOH (in liters)

Rearranging the equation to solve for M1:

M1 = (M2 * V2) / V1

Substituting the given values:

M1 = (0.20 M * 0.023 L) / 0.030 L

Calculating:

M1 ≈ 0.153 M

So, the molarity of HCl in the solution is approximately 0.153 M.