Chemical Reactions

3.a. $Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)$

3.b.

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3.c. Moles of $CaCO_3$ = $\frac{0.93}{100.1}$ = 0.0093 mol

Since the mole ratio of $Na_2CO_3$ to $CaCO_3$ is 1:1, the moles of $Na_2CO_3$ = 0.0093 mol.

3.d. The student’s claim is incorrect. If the precipitate was not completely dried, the mass of the precipitate would be too high, resulting in a calculated molarity that is too high.

3.e. The solution would conduct electricity because there are mobile ions present in the solution ($Na^+$, $NO_3^-$).

3.f.i. A pH meter can be used to measure the pH of the solution. The pOH can be calculated using the equation pOH = 14 - pH, and then $[OH^-]$ can be calculated using the equation $[OH^-]$ = $10^{-pOH}$.

3.f.ii. The $[OH^-]$ can be used to calculate $[HCO_3^-]$ using the equilibrium expression. Because the stoichiometry of the reaction is 1:1, $[OH^-]$ = $[HCO_3^-]$. The initial concentration of $CO_3^{2-}$ can be calculated using the equilibrium expression and the ICE table.

3.g. The concentration of $HCO_3^-(aq)$ is less than the concentration of $CO_3^{2-}(aq)$. Since $K_b$ is less than 1, the equilibrium favors the reactants.

3.h. No, the $Na_2CO_3$ solution is not suitable for making a buffer with a pH of 6. The $CO_3^{2-}/HCO_3^-$ buffer system will have a basic pH because $CO_3^{2-}$ is a base.

7.a. $MnO_4^-$ is reduced. The oxidation number of Mn changes from +7 in $MnO_4^-$ to +2 in $Mn^{2+}$.

7.b. Initial buret reading = 1.50 mL

Final buret reading = 26.85 mL

Volume of $KMnO_4$ added = 26.85 - 1.50 = 25.35 mL

7.c. Moles of $MnO_4^-$ = 0.02535 L * 0.0235 mol/L = 0.000596 mol

7.d. No, this concentration is not a reasonable choice. The volume of titrant required to reach the endpoint would be too large, which would lead to a large percent error in the measurement. The titrant would also be added too quickly, making it difficult to stop at the endpoint.