Q25P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

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

Step 1: 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.

Step 2: 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 $\mathrm{pH} < 7$.

Step 3: 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 $\mathrm{pOH} < 7$ .