AP Chemistry 8.8 Properties of Buffers - Exam Style questions - FRQs- New Syllabus

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

\( \mathrm{Na^+} \) and \( \mathrm{NO_3^-} \) are spectator ions, so they are not included in the net ionic equation.

(b)
The completed particulate diagram should show one \( \mathrm{Ca^{2+}} \) ion still present in the solution, along with the \( \mathrm{Na^+} \) and \( \mathrm{NO_3^-} \) ions already shown.

Because \( \mathrm{Ca(NO_3)_2} \) was added in excess, some \( \mathrm{Ca^{2+}} \) remains in the solution after all of the \( \mathrm{CO_3^{2-}} \) has precipitated as \( \mathrm{CaCO_3}(s) \).

(c)
First calculate moles of \( \mathrm{CaCO_3} \):

\( n = \dfrac{m}{M} = \dfrac{0.93\ \mathrm{g}}{100.1\ \mathrm{g/mol}} = 9.29 \times 10^{-3}\ \mathrm{mol} \)

Since the reaction ratio is \( 1:1 \), \( \mathrm{Na_2CO_3} : \mathrm{CaCO_3} = 1:1 \), the moles of \( \mathrm{Na_2CO_3} \) originally present are also \( 9.29 \times 10^{-3}\ \mathrm{mol} \).

To appropriate significant figures: \( 0.0093\ \mathrm{mol\ Na_2CO_3} \)

(d)
No, I do not agree.

If the precipitate was not completely dry, the measured mass would be greater than the actual mass of pure \( \mathrm{CaCO_3} \), because some of the mass would come from water.

That would make the calculated moles of \( \mathrm{CaCO_3} \) too large, which would also make the calculated moles and molarity of \( \mathrm{Na_2CO_3} \) too high, not too low.

(e)
The liquid would conduct electricity.

Even after the precipitate is removed, dissolved ions remain in the filtrate, including \( \mathrm{Na^+}(aq) \), \( \mathrm{NO_3^-}(aq) \), and excess \( \mathrm{Ca^{2+}}(aq) \). These mobile ions allow the solution to conduct an electric current.

(f)(i)
The student could determine the \( \mathrm{pH} \) of the solution using a \( \mathrm{pH} \) meter.

(f)(ii)
First, use the measured \( \mathrm{pH} \) to find \( \mathrm{pOH} \):

\( \mathrm{pOH} = 14 – \mathrm{pH} \)

Then determine \( [\mathrm{OH^-}] \):

\( [\mathrm{OH^-}] = 10^{-\mathrm{pOH}} \)

Next, use an ICE table for \( \mathrm{CO_3^{2-}}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{HCO_3^-}(aq) + \mathrm{OH^-}(aq) \). If the equilibrium concentration of \( \mathrm{OH^-} \) is \( x \), then the equilibrium concentration of \( \mathrm{HCO_3^-} \) is also \( x \), and the equilibrium concentration of \( \mathrm{CO_3^{2-}} \) is \( c_i – x \), where \( c_i \) is the initial concentration of \( \mathrm{CO_3^{2-}} \).

Substitute into the expression for \( K_b \):

\( K_b = \dfrac{[\mathrm{HCO_3^-}][\mathrm{OH^-}]}{[\mathrm{CO_3^{2-}}]} = \dfrac{(x)(x)}{c_i – x} \)

Then solve for \( c_i \):

\( c_i = \dfrac{x^2}{K_b} + x \)

Thus, using the measured \( \mathrm{pH} \) value to find \( x = [\mathrm{OH^-}] \), the student can determine the initial concentration of \( \mathrm{CO_3^{2-}} \).

(g)
The concentration of \( \mathrm{HCO_3^-}(aq) \) is less than the concentration of \( \mathrm{CO_3^{2-}}(aq) \).

Because \( K_b = 2.1 \times 10^{-4} \) is small, the hydrolysis reaction is reactant-favored. Therefore, most of the dissolved species remains as \( \mathrm{CO_3^{2-}} \), and only a smaller amount is converted to \( \mathrm{HCO_3^-} \).

(h)
No, the \( \mathrm{Na_2CO_3} \) solution is not suitable for making a buffer with \( \mathrm{pH} = 6 \).

A buffer works best when the desired \( \mathrm{pH} \) is close to the \( \mathrm{p}K_a \) of the weak acid in the conjugate acid-base pair. Here, the weak acid is \( \mathrm{HCO_3^-} \), and its \( \mathrm{p}K_a \) is approximately \( 10.32 \).

Since \( 6 \) is far from \( 10.32 \), the \( \mathrm{CO_3^{2-}/HCO_3^-} \) pair would not function as an effective buffer at \( \mathrm{pH} = 6 \).