Question

Question: Derive an equation to solve for the pH for a dilute weak base in which water is a significant contributor to the pH. Use this formula to calculate the pH of a 100.0-mL sample of 2.0x10^-5 MB (where B is a weak base with a Kb value of 6.1x10^-11).

Short answer

Step 1: Derive equation for weak base BThe pH of the solution is 7.04.

Step-by-step solution

Derive equation for weak base B

The K_b for weak base B is,

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

Derivation of equation,

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

Derivation of equation for [B],

$$\begin{gathered} {\left[B\right]}_0=\left[B\right]+\left[HB\right] \\ {\left[B\right]}_0=\left[B\right]+\left[O{H}^{-}\right]-\frac{K_w}{\left[O{H}^{-}\right]} \\ \left[B\right]={\left[B\right]}_0-\left(\left[O{H}^{-}\right]-\frac{K_w}{\left[O{H}^{-}\right]}\right) \end{gathered}$$

Substitute above equations in K_b equation,

$$\begin{gathered} K_b=\frac{\left[O{H}^{-}\right]\left(\left[O{H}^{-}\right]-\frac{K_w}{\left[O{H}^{-}\right]}\right)}{{\left[B\right]}_0-\left(\left[O{H}^{-}\right]-\frac{K_w}{\left[O{H}^{-}\right]}\right)} \\ {K}_b=\frac{\left[O{H}^{-}\right]-K_w}{{\left[B\right]}_0-\left(\frac{{\left[O{H}^{-}\right]}^2-K_w}{\left[O{H}^{-}\right]}\right)} \end{gathered}$$

 Step 2: Calculation of [OH-]

The value of [OH^-] is,

$$\begin{gathered} 6.1\times {10}^{-11}=\frac{x.x}{2.0\times {10}^{-5}-x} \\ x=3.5\times {10}^{-8} \end{gathered}$$

Calculation of [OH^-] using derived equation,

$$\begin{gathered} 6.1\times {10}^{-11}=\frac{{\left[O{H}^{-}\right]}^2-1.0\times {10}^{-14}}{2.0\times {10}^{-5}-\left(\frac{{\left[1.0\times {10}^{-7}\right]}^2-1.0\times {10}^{-14}}{1.0\times {10}^{-7}}\right)} \\ \left[O{H}^{-}\right]=1.1\times {10}^{-7} \end{gathered}$$

 Step 3: Determination of pH

The pH of the solution is,

$$\begin{gathered} OH=-\log \left[O{H}^{-}\right] \\ pOH=-\log \left(1.1\times {10}^{-7}\right) \\ pOH=6.96 \\ pH=14.0-6.96 \\ pH=7.04 \end{gathered}$$

Thus, the pH of the solution is 7.04.