Question

The $\mathit{p}{\mathit{K}}_{\mathit{k}}$of seawater at $\mathbf{25}\mathbf{°}\mathit{C}$is 13.776. This differs from the usual$\mathit{p}{\mathit{K}}_{\mathit{k}}$Of 14.00 at this temperature because dissolved salts make sea water a non-ideal solution. If the pH is seawater is 8.00, water are the concentration of ${\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}$and $\mathit{O}{\mathit{H}}^{\mathbf{-}}$ in seawater at $\mathbf{25}\mathbf{°}\mathit{C}$.

Short answer

The concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ion will be$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]==1.0\times {10}^{-8}$ , and the concentration of $\left[{\mathrm{OH}}^{-}\right]$ion will be$\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-6}$ at $25°\mathrm{C}$.

Step-by-step solution

Understanding the concept of pH of an acid-base.

The negative logarithm of ${\mathrm{H}}^{+}$ ion concentration is used to calculate pH. It's a scale that is used to measure acids and bases.

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

Also we know that$\mathrm{pOH}+\mathrm{pH}=14$

Applying the concept and calculation of concentration of hydronium ion.

Given value of pH is 8.

Now to calculate the range of concentration of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ ion,first calculate the concentration at the pH value of 8

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ 8=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-8} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-8}\mathrm{M} \end{gathered}$$

Calculation of Kw value.

Given pKw value of sea water is 13.776.