Question

If 56.0mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 740mg sample of Na2SO4, what is the molarity of the solution?

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If 43.0mL of 0.210 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution?

Answer 1

Answer on Question#39839 - Chemistry - Other

Questions

1) If 56.0 mL of BaCl₂ solution is needed to precipitate all the sulfate ion in a 740 mg sample of Na₂SO₄, what is the molarity of the solution?

2) If 43.0 mL of 0.210 M HCl solution is needed to neutralize a solution of Ca(OH)₂, how many grams of Ca(OH)₂ must be in the solution?

Answer

1) The total chemical reaction of this process:

$$\mathrm{Na_2SO_4} + \mathrm{BaCl_2} = 2\mathrm{NaCl} + \mathrm{BaSO_4}$$

The molarity of a solution could be calculated according to the formula:

$$C_M = \frac{v}{V}$$

where v - moles of the solute, moles; V - volume of the solvent, l.

$$v = \frac{m}{M}$$

where m - mass of the solute, grams; M - molar mass of the solute, gram/moles.

$$\begin{array}{l} M(\mathrm{Na_2SO_4}) = 142\ \mathrm{g/mol} \\ v(\mathrm{Na_2SO_4}) = \frac{0.74}{142} = 0.0052\ \mathrm{mol} \\ v(\mathrm{Na_2SO_4}) = v(\mathrm{BaCl_2}) = 0.0052\ \mathrm{mol} \\ C_M(\mathrm{BaCl_2}) = \frac{0.0052}{0.056} = 0.093\ \mathrm{M} \\ \end{array}$$

2) The total chemical reaction of this process:

$$2\mathrm{HCl} + \mathrm{Ca(OH)_2} = \mathrm{CaCl_2} + 2\mathrm{H_2O}$$

The molarity of a solution could be calculated according to the formula:

$$C _ {M} = \frac {v}{V}$$

where v- moles of the solute, moles; V-volume of the solvent, l.

According to this equation, the amount of moles is:

$$v = C _ {M} \cdot V$$ $$v (HCl) = 0.210 \cdot 0.043 = 0.0090 \text{ mol}$$ $$v \left(\mathrm{Ca(OH)}_2\right) = \frac {v (HCl)}{2} = \frac {0.0090}{2} = 0.0045 \text{ mol}$$ $$v = \frac {m}{M} \quad m = v \cdot M$$

where m-mass of the solute, grams; M-molar mass of the solute, gram/moles.

$$M \left(\mathrm{Ca(OH)}_2\right) = 74 \text{ g/mol}$$ $$m \left(\mathrm{Ca(OH)}_2\right) = 0.00452 \cdot 74 = 0.33 \text{ g}$$