Question

Sodium acetate hydrolysis treated by Solver with activity coefficients.

(a) Following the NH₃ example in Section 8-5, write the equilibria and charge and mass balances needed to find the composition of 0.01 M sodium acetate (Na⁺A^-). Include activity coefficients where appropriate. The two reactions are hydrolysis (pK_b = 9.244) and ionization of H₂O.

(b) Including activity coefficients, set up a spreadsheet analogous to Figure 8-12 to find the concentrations of all species. Assign an initial value of ionic strength = 0.01. After the rest of the spreadsheet is set up, change the ionic strength from the numerical value 0.01 to the correct formula for ionic strength. This two-step process of beginning with a numerical value and then going to a formula is necessary because of circular references between ionic strength and concentrations that depend on ionic strength. There are four unknowns and two equilibria, so use Solver to find 4 - 2 = 2 concentrations (pC values). Solver does not find both pC values at the same time well in this problem. Execute one pass to find both pC values by varying pA and pOH to minimize$\mathit{\Sigma }{{\mathit{b}}_{\mathbf{i}}}^{\mathbf{2}}$ . Then vary only pA to minimize$\mathit{\Sigma }{{\mathit{b}}_{\mathbf{i}}}^{\mathbf{2}}$ . Then vary only pOH to minimize $\mathit{\Sigma }{{\mathit{b}}_{\mathbf{i}}}^{\mathbf{2}}$. Continue alternating to solve for one value at a time as long as $\mathit{\Sigma }{{\mathit{b}}_{\mathbf{i}}}^{\mathbf{2}}$ continues to decrease. Find [A^-], [OH^-], [HA], and [H⁺]. Find the ionic strength, pH =-log([H⁺] ${\mathit{\gamma }}^{\mathbf{+}}$) and the fraction of hydrolysis = [HA]/F.

Short answer

(a)

The equilibrium reactions are

$\begin{array}{rcl}& & A^{-}+H_{2}O\stackrel{K_b}{\rightleftharpoons }HA+O{H}^{-}\\ & & H_{2}O\stackrel{K_w}{\rightleftharpoons }{H}^{+}+O{H}^{-}\end{array}$

Charge balance is written as follows

$\left[H^{+}\right]+\left[N{a}^{+}\right]=\left[O{H}^{-}\right]+\left[A^{-}\right]$

Mass balance is written as follows

$\begin{array}{rcl}\left[N{a}^{+}\right]& =& 0.01M=F\\ \left[HA\right]+\left[A^{-}\right]& =& 0.01M=F\end{array}$

(b)

The values obtained are as follows

[A^-]=1.00×10^-2, [OH^-]=2.40×10^-06, [HA]= 2.40×10^-06, and [H⁺]=4.20×10^-09

Ionic strength of H⁺ is 4.20×10^-09M

The pH value is 8.4

Fraction of hydrolysis=HA/F=2.4×10^-04

Step-by-step solution

Calculation regarding part (a)

(a)

The pertinent reactions are

$\begin{array}{rcl}& & A^{-}+H_{2}O\stackrel{K_b}{\rightleftharpoons }HA+O{H}^{-}----\left(1\right)\\ & & H_{2}O\stackrel{K_w}{\rightleftharpoons }{H}^{+}+O{H}^{-}------\left(2\right)\\ & & Given,\\ p{K}_b& =& 9.244\\ -{\log }_{10}{K}_b& =& -9.244\\ K_b& =& 5.7\times {10}^{-10}\\ K_w& =& 1.0\times {10}^{-14}\\ K_b& =& \frac{\left[HA\right]{\gamma }_{HA}\left[O{H}^{-}\right]{\gamma }_{O{H}^{-}}}{\left[A^{-}\right]{\gamma }_{A^{-}}}\\ K_w& =& \left[H^{+}\right]{\gamma }_{H^{+}}\left[O{H}^{-}\right]{\gamma }_{O{H}^{-}}\end{array}$

Charge balance is written as follows

$\left[H^{+}\right]+\left[N{a}^{+}\right]=\left[O{H}^{-}\right]+\left[A^{-}\right]-----\left(3\right)$

Mass balance is written as follows

$\begin{array}{rcl}\left[N{a}^{+}\right]& =& 0.01M=F----\left(4\right)\\ \left[HA\right]+\left[A^{-}\right]& =& 0.01M=F---\left(5\right)\end{array}$

Calculation regarding part (b)

(b)

Neglecting the activity coefficient, the following changes can be substituted in charge balance

$\left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]}$

From equation (1) :$\left[HA\right]=\frac{K_b\left[A^{-}\right]}{\left[O{H}^{-}\right]}------\left(6\right)$

From equation (5) :$HA=F-\left[A\right]---\left(7\right)$

Now equating the expressions of (6) and (7)

$\begin{array}{rcl}\frac{K_b\left[A^{-}\right]}{\left[O{H}^{-}\right]}& =& F-\left[A^{-}\right]\\ \left[A^{-}\right]\left(\frac{K_b}{\left[O{H}^{-}\right]}+1\right)& =& F\\ \left[A^{-}\right]& =& \frac{F\left[O{H}^{-}\right]}{K_b+\left[O{H}^{-}\right]}\\ \left[A^{-}\right]& =& \frac{F\left[K_w\right]}{K_b{H}^{+}+\left[K_w\right]}substitutingO{H}^{-}-----\left(8\right)\\ & & \\ & & \end{array}$

Charge balance

$\begin{array}{rcl}\left[H^{+}\right]+\left[N{a}^{+}\right]& =& \left[O{H}^{-}\right]+\left[A^{-}\right]\\ \left[\frac{K_w}{\left[H^{+}\right]}\right]+\frac{F{K}_w}{K_b\left[H^{+}\right]+K_w}-\left[H^{+}\right]-\left[F\right]& =& 0------\left(9\right)\end{array}$

Using SOLVER the following spreadsheet was obtained

The values obtained are as follows

[A^-]=1.00×10^-2, [OH^-]=2.40×10^-06, [HA]= 2.40×10^-06, and [H⁺]=4.20×10^-09

Ionic strength of H⁺ is 4.20×10^-09M

The pH value

pH =-log([H⁺] )=8.4

Fraction of hydrolysis=HA/F=2.4×10^-04