Question

20mL of Al(OH)3 reacts completely with 50mL of 1mol of HBr. What is the concentration of Al(OH)3?

Answer 1

20mL of $\mathrm{Al(OH)_3}$ reacts completely with 50mL of 1mol of HBr. What is the concentration of $\mathrm{Al(OH)_3}$?

Solution:

We write the equation for the reaction:

$$\mathrm{Al(OH)_3} + 3\mathrm{HBr} = \mathrm{AlBr_3} + 3\mathrm{H_2O}$$ $$M(\mathrm{Al(OH)_3}) = \frac{n(\mathrm{Al(OH)_3})}{v(\mathrm{Al(OH)_3})} = \frac{\text{mole}}{l}$$ $$M - \text{concentration of the substance}$$

Convert your mL solutions to L in order to plug them into formula.

$$v\left((\mathrm{Al(OH)_3})\right) = 20\mathrm{ml} \times \frac{1\mathrm{l}}{1000\mathrm{ml}} = 0.020\mathrm{l}$$ $$v(\mathrm{HBr}) = 50\mathrm{ml} \times \frac{1\mathrm{l}}{1000\mathrm{ml}} = 0.050\mathrm{l}$$

Find the amount of the substance HBr, which came in response:

$$n(\mathrm{HBr}) = V(\mathrm{HBr}) \times M(\mathrm{HBr})$$ $$n(\mathrm{HBr}) = 1 \times 0.050 = 0.05\ \text{mole}$$

According to the reaction equation we find the amount of the substance $\mathrm{Al(OH)_3}$:

$$\frac{n(\mathrm{Al(OH)_3})}{n(\mathrm{HBr})} = \frac{1}{3}$$ $$n(\mathrm{Al(OH)_3}) = \frac{n(\mathrm{HBr})}{3} = \frac{0.05}{3} = 0.017\ \text{mole}$$

Find the concentration of $\mathrm{Al(OH)_3}$:

$$M(\mathrm{Al(OH)_3}) = \frac{0.017\ \text{mole}}{0.020\ \mathrm{l}} = 0.85$$

Answer: 0.85 M $(\mathrm{Al(OH)_3})$