Question

Use Table 6-1 to find the $\left[H^{+}\right]$in water at $\mathbf{100}\mathbf{°}\mathbf{C}$and at$\mathbf{0}\mathbf{°}\mathbf{C}$.

Short answer

$\left[{\mathrm{H}}^{+}\right]$in the water at $100°C$ and at $0°C$are found to be $7.4\times {10}^{-7}$ and ,$3.4\times {10}^{-8}$ respectively.

Step-by-step solution

Define the formula for finding Kb.

The value of can be calculated as:

${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\left[{\mathrm{BH}}^{+}\right]\left[\mathrm{OH}\right]}{\left[B\right]}$

Where ${\mathbf{K}}_{\mathbf{b}}$is called the base dissociation constant.

Step 2: Calculate the Kw.

Given:

At${100}^{°}C,K_w=5.43\times {10}^{-13}$

At$0^{°}C{K}_w=1.15\times {10}^{-15}$

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

$\left[{\mathrm{H}}^{+}\right]$in the water at$100°C$

$$\begin{gathered} K_w=5.43\times {10}^{-14} \\ =\left[H^{+}\right]\left[O{H}^{-}\right] \\ =\left[x\right]\left[x\right]=[x{]}^{2}x \\ =5.43\times {10}^{-14} \\ =7.4\times {10}^{-7} \end{gathered}$$

At $0°C$

$$\begin{gathered} K_w=1.15\times {10}^{-15} \\ =\left[H^{+}\right]\left[O{H}^{-}\right] \\ =\left[x\right]\left[x\right]=[x{]}^{2}x \\ =\sqrt{1.15\times {10}^{-15}} \\ =3.4\times {10}^{-8} \end{gathered}$$

Using temperature dependent of $K_w^a$values $\left[H^{+}\right]$in the water.