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a. If 150 mL of [tex]$Al(OH)_3$[/tex] of an unknown concentration with a methyl red indicator present is titrated with [tex]$0.535 M$[/tex] [tex][tex]$HCl$[/tex][/tex], the yellow solution turns red after 32.8 mL of the [tex]$HCl$[/tex] is added. What is the molarity of the [tex]$Al(OH)_3$[/tex] solution?

Sagot :

GinnyAnsweridn
To find the molarity of the [tex]\( \text{Al(OH)}_3 \)[/tex] solution, we need to use the data from the titration process. Here is a step-by-step solution:

1. Identify the given values:
- Volume of [tex]\( \text{Al(OH)}_3 \)[/tex] solution, [tex]\( V_{\text{Al(OH)}_3} \)[/tex] = 150 mL
- Concentration of [tex]\( \text{HCl} \)[/tex] solution, [tex]\( C_{\text{HCl}} \)[/tex] = 0.535 M
- Volume of [tex]\( \text{HCl} \)[/tex] used, [tex]\( V_{\text{HCl}} \)[/tex] = 32.8 mL

2. Convert volumes from milliliters to liters, since molarity is expressed in moles per liter:
- [tex]\( V_{\text{Al(OH)}_3} \)[/tex] in liters: [tex]\( 150 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.150 \, \text{L} \)[/tex]
- [tex]\( V_{\text{HCl}} \)[/tex] in liters: [tex]\( 32.8 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0328 \, \text{L} \)[/tex]

3. Calculate the number of moles of [tex]\( \text{HCl} \)[/tex]:
- Number of moles of [tex]\( \text{HCl} \)[/tex] = [tex]\( C_{\text{HCl}} \times V_{\text{HCl}} = 0.535 \, \text{M} \times 0.0328 \, \text{L} = 0.017548 \, \text{moles} \)[/tex]

4. Write the balanced chemical equation for the reaction:
[tex]\[ \text{Al(OH)}_3 + 3\text{HCl} \rightarrow \text{AlCl}_3 + 3\text{H}_2\text{O} \][/tex]
This equation tells us that 1 mole of [tex]\( \text{Al(OH)}_3 \)[/tex] reacts with 3 moles of [tex]\( \text{HCl} \)[/tex].

5. Calculate the number of moles of [tex]\( \text{Al(OH)}_3 \)[/tex] that reacted using the stoichiometry from the balanced equation:
- Moles of [tex]\( \text{Al(OH)}_3 \)[/tex] = [tex]\( \frac{\text{Moles of HCl}}{3} = \frac{0.017548 \, \text{moles}}{3} = 0.005849 \, \text{moles} \)[/tex]

6. Finally, calculate the molarity of the [tex]\( \text{Al(OH)}_3 \)[/tex] solution:
- Molarity ([tex]\( C_{\text{Al(OH)}_3} \)[/tex]) = [tex]\( \frac{\text{Moles of Al(OH)}_3}{V_{\text{Al(OH)}_3}} = \frac{0.005849 \, \text{moles}}{0.150 \, \text{L}} = 0.038996 \, \text{M} \)[/tex]

So, the molarity of the [tex]\( \text{Al(OH)}_3 \)[/tex] solution is approximately [tex]\( 0.038996 \, \text{M} \)[/tex].
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