A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

Step by step solution

01

Calculate moles of \(\mathrm{CaCO}_3\)

First, calculate the moles of \(\mathrm{CaCO}_3\) from the given mass. The molar mass of \(\mathrm{CaCO}_3\) is 40.08 + 12.01 + (3 * 16.00) = 100.09 g/mol. So, there are \(0.1000 \mathrm{g} / 100.09 \mathrm{ g/mol} = 0.001 \mathrm{ mol}\) of \(\mathrm{CaCO}_3\)

02

Calculate moles of \(\mathrm{HCl}\) reacted with \(\mathrm{CaCO}_3\)

From the balanced chemical equation, \(\mathrm{CaCO}_3\) reacts with \(\mathrm{HCl}\) in a 1:2 ratio. Therefore, the moles of \(\mathrm{HCl}\) reacted with \(\mathrm{CaCO}_3\) are \(2 * 0.001 \mathrm{moles} = 0.002 \mathrm{ mol}\)

03

Calculate moles of excess \(\mathrm{HCl}\) titrated

To find the moles of excess \(\mathrm{HCl}\), use the volume and molarity of \(\mathrm{Ba(OH)}_2\) used in titration. The moles of \(\mathrm{Ba(OH)}_2\) are \(43.82 \mathrm{mL} * 0.01185 \mathrm{M} = 0.000519 \mathrm{ mol}\). From the balanced chemical reaction, \(\mathrm{Ba(OH)}_2\) reacts with \(\mathrm{HCl}\) in a 1:2 ratio, so the moles of \(\mathrm{HCl}\) are \(2 * 0.000519 \mathrm{ mol} = 0.001038 \mathrm{ mol}\)

04

Calculate total moles of \(\mathrm{HCl}\) and molarity

The total moles of \(\mathrm{HCl}\) are the sum of the \(\mathrm{HCl}\) that reacted with \(\mathrm{CaCO}_3\) and the excess \(\mathrm{HCl}\). The total moles = \(0.002 \mathrm{ mol} + 0.001038 \mathrm{ mol} = 0.003038 \mathrm{ mol}\). The molarity is the total moles divided by the total volume in liters, which is \((25.00 \mathrm{mL} + 43.82 \mathrm{mL}) / 1000 = 0.06882 \mathrm{L}\). So, the molarity of the original \(\mathrm{HCl}\) is \(0.003038 \mathrm{ mol} / 0.06882 \mathrm{ L} = 0.0441 \mathrm{ M}\)

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