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In a different titration experiment, it took 19.26 mL of a 0.162 M solution of NaOH to titrate 20.00 mL of an unknown HCl solution. Calculate the molarity of the HCl solution.

[tex]\[ \text{HCl (aq) + NaOH (aq) } \longrightarrow \text{ NaCl (aq) + H}_2\text{O (l)} \][/tex]


Sagot :

To determine the molarity of the unknown HCl solution, we need to follow these steps systematically:

1. Identify the volume and molarity of the NaOH solution used:
- Volume of NaOH solution ([tex]\( V_{\text{NaOH}} \)[/tex]) = 19.26 mL
- Molarity of NaOH solution ([tex]\( M_{\text{NaOH}} \)[/tex]) = 0.162 M

2. Calculate the moles of NaOH used in the reaction:
To find the moles of NaOH, we use the formula for moles:
[tex]\[ \text{Moles of NaOH} = \text{Molarity of NaOH} \times \text{Volume of NaOH (in liters)} \][/tex]
Since the volume is given in milliliters, we convert it to liters by dividing by 1000:
[tex]\[ \text{Volume of NaOH in liters} = \frac{19.26 \ \text{mL}}{1000} = 0.01926 \ \text{L} \][/tex]
Now, calculate the moles of NaOH:
[tex]\[ \text{Moles of NaOH} = 0.162 \ \text{M} \times 0.01926 \ \text{L} = 0.00312012 \ \text{moles} \][/tex]

3. Determine the moles of HCl reacted:
The balanced chemical equation [tex]\( \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \)[/tex] shows that the reaction occurs in a 1:1 molar ratio. Hence, the moles of HCl that reacted will be equal to the moles of NaOH used:
[tex]\[ \text{Moles of HCl} = \text{Moles of NaOH} = 0.00312012 \ \text{moles} \][/tex]

4. Identify the volume of the HCl solution used:
- Volume of HCl solution ([tex]\( V_{\text{HCl}} \)[/tex]) = 20.00 mL

5. Calculate the molarity of the HCl solution:
To find the molarity of the HCl solution, we use the formula for molarity:
[tex]\[ \text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl (in liters)}} \][/tex]
Convert the volume of HCl to liters:
[tex]\[ \text{Volume of HCl in liters} = \frac{20.00 \ \text{mL}}{1000} = 0.02000 \ \text{L} \][/tex]
Now, calculate the molarity of the HCl solution:
[tex]\[ \text{Molarity of HCl} = \frac{0.00312012 \ \text{moles}}{0.02000 \ \text{L}} = 0.156006 \ \text{M} \][/tex]

Therefore, the molarity of the unknown HCl solution is [tex]\(0.156006 \ \text{M}\)[/tex].