The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.2 mL of the NaOH solution to reach the end point of the titration. A buret filled with a titrant is held above a graduated cylinder containing an analyte solution. What is the initial concentration of HCl

The unit of concentration [tex]\rm M[/tex] is the same as [tex]\rm mol \cdot L^{-1}[/tex] (moles per liter.) On the other hand, the volume of both the [tex]\rm NaOH[/tex] solution and the original [tex]\rm HCl[/tex] solution here are in milliliters. Convert these two volumes to liters:

[tex]V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L[/tex].
[tex]V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L[/tex].

Calculate the number of moles of [tex]\rm NaOH[/tex] in that [tex]0.0182\; \rm L[/tex] of [tex]0.160\; \rm M[/tex] solution:

[tex]\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}[/tex].

[tex]\rm HCl[/tex] reacts with [tex]\rm NaOH[/tex] at a one-to-one ratio:

[tex]\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)[/tex].

Coefficient ratio:

[tex]\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1[/tex].

In other words, one mole of [tex]\rm NaOH[/tex] would neutralize exactly one mole of [tex]\rm HCl[/tex]. In this titration, [tex]0.291\; \rm mol[/tex] of [tex]\rm NaOH\![/tex] was required. Therefore, the same amount of [tex]\rm HC[/tex] should be present in the original solution:

[tex]\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}[/tex].

Calculate the concentration of the original [tex]\rm HCl[/tex] solution:

[tex]\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M[/tex].