Question

(a) What is the pH of $\mathbf{6}\mathbf{.}\mathbf{14}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$M HI? Is the solution neutral, acidic, or basic?

(b) What is the pOH of 2.55 M Ba ${\left(\mathrm{OH}\right)}_{\mathbf{2}}$? Is the solution neutral, acidic, or basic?

Short answer

(a) The solution $6.14\times {10}^{-3}$MHIis acidic.

(b) The solution$2.55/\mathrm{M}/\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ is highly basic.

Step-by-step solution

Concept Introduction

An acid is a type of chemical that produces sour solutions due to a high concentration of positive hydrogen ions in the solution.

A base is a type of chemical that produces bitter-tasting solutions due to a lack of positive hydrogen ions. Acids and bases combine to generate salts.

The term "neutral" refers to a substance that is in the middle of the acid-base spectrum.

Is the solution neutral, acidic, or basic?

(a)

$6.14\times {10}^{-3}\mathrm{MHI}$:

Let's start with the dissociation of strong acid HI in water.

$\mathrm{HI}\to {\mathrm{H}}^{+}+{\mathrm{I}}^{-}$

The concentration of HI is equal to the concentration of both $\left[{\mathrm{H}}^{+}\right]$and $\left[{\mathrm{I}}^{-}\right]$.

From this information, we can solve for the $\mathrm{pH}$.

$$\begin{gathered} \mathrm{pH}=‐\log \left[{\mathrm{H}}^{+}\right] \\ =‐\log \left(6.14\times {10}^{-3}\mathrm{M}\right) \\ \mathrm{pH}=2.21 \end{gathered}$$

Therefore, the solution is acidic because of the $\mathrm{pH}<7$.

Is the solution neutral, acidic, or basic?

(b)

$2.55/\mathrm{M}/\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$:

Let's start with the strong base $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$dissociation reaction in water.

$\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\to {\mathrm{Ba}}^{2+}+2{\mathrm{OH}}^{-}$

The concentration of $\mathrm{Ba}{\left(\mathrm{OH}\right)}_2$is equal to the concentration of $\left[{\mathrm{Ba}}^{2+}\right]$and$\left[{\mathrm{OH}}^{-}\right]=2\left[\mathrm{Ba}{\left(\mathrm{OH}\right)}_2\right]$.

So

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=2\left(2.55\mathrm{M}\right) \\ =5.10\mathrm{M} \end{gathered}$$

From this information, we can solve for the pOH.

$$\begin{gathered} \mathrm{pOH}=‐\log \left[{\mathrm{OH}}^{+}\right] \\ =‐\log \left(5.10\mathrm{M}\right) \\ \mathrm{pOH}=‐0.707 \end{gathered}$$

Therefore, the answer is really basic because $\mathrm{pOH}<7$.