Question

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

Answer 1

Approximately
`0.291\; \rm M` (rounded to two significant figures.)

Step-by-step explanation:

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

• 
  `V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 * 10^(-3)\; \rm L = 0.0182\; \rm L`.
• 
  `V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0* 10^(-3)\; \rm L = 0.0100\; \rm L`.

Calculate the number of moles of
`\rm NaOH` in that
`0.0182\; \rm L` of
`0.160\; \rm M` solution:

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

`\rm HCl` reacts with
`\rm NaOH` at a one-to-one ratio:

`\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)`.

Coefficient ratio:

`\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1`.

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

`\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol * 1 = 0.00291\; \rm mol\end{aligned}`.

Calculate the concentration of the original
`\rm HCl` solution:

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