Q. 10.29 - An Introduction to General, Organic, and Biological Chemistry

Question

Indicate whether each of the following solutions is acidic, basic, or neutral:

a. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=2.0\times {10}^{-5}\mathrm{M}$

b. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.4\times {10}^{-9}\mathrm{M}$

c. $\left[{\mathrm{OH}}^{-}\right]=8.0\times {10}^{-3}\mathrm{M}$

d. $\left[{\mathrm{OH}}^{-}\right]=3.5\times {10}^{-10}\mathrm{M}$

Step-by-Step Solution

Answer

a. The solution is acidic nature.

b. The solution is basic nature.

c. The solution is basic nature.

d. The solution is acidic nature.

Step 1: Introduction

To answer the questions, use the ionic product constant of water.

The term representing water's ionic product is,

$K_{\mathrm{W}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

To generate pure water,

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.0\times {10}^{-7}\mathrm{M}$

$\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-7}\mathrm{M}$

Then,

$K_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

$=\left(1.0\times {10}^{-7}\mathrm{M}\right)\times \left(1.0\times {10}^{-7}\mathrm{M}\right)$

$K_{\mathrm{W}}=1.0\times {10}^{-14}\mathrm{M}$

Step 2: Explanation

At $25°\mathrm{C}$, this quantity is constant for the aby aqueous medium.

If $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$, the solution is neutral. $\left[{\mathrm{OH}}^{-}\right]$and are the same.

When acid is introduced to water, the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ equation is modified. increases and $\left[{\mathrm{OH}}^{-}\right]$increases decreases. If $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ is true,  is more than $\left[{\mathrm{OH}}^{-}\right]$ The solution is acidic in nature.

$\left[{\mathrm{OH}}^{-}\right]$ is more than $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$. The answer  and the solution is simple.

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]>\left[{\mathrm{OH}}^{-}\right]\to Acidic$

$\left[{\mathrm{OH}}^{-}\right]>\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\to Basic$

Step 3: Given information part(a)

To find whether the given solutions is acidic, basic, or neutral.

Step 4: Explanation part(a)

Because of this, the concentration of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$as $2.0\times {10}^{-5}\mathrm{M}$.

Find the concentration of ${\mathrm{OH}}^{-}$ by substituting the values of the ionic product of water and the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ in the formula of the ionic product of water.

$K_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

$\left[{\mathrm{OH}}^{-}\right]=\frac{K_{\mathrm{W}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}$

$\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{2.0\times {10}^{-5}\mathrm{M}}$

$\left[{\mathrm{OH}}^{-}\right]=5.0\times {10}^{-10}\mathrm{M}$

Step 5: Explanation part(a)

Compare the values of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ and $\left[{\mathrm{OH}}^{-}\right]$, in this case, the value of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$this indicates that the fluid is acidic.

$2.0\times {10}^{-5}>5.0\times {10}^{-10}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]>\left[{\mathrm{OH}}^{-}\right]\to Acidic$

The proportion of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ in acidic solution is higher.

As a result, the solution has an acidic $pH$.

Step 6: Given information part(b)

To find whether the given solutions is acidic, basic, or neutral.

Step 7: Explanation part(b)

On account of this, the concentration of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$as $1.4\times {10}^{-9}\mathrm{M}$.

Find the concentration of ${\mathrm{OH}}^{-}$ by substituting the values of the ionic product of water and the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ in the formulation of the ionic product of water.

$K_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

$\left[{\mathrm{OH}}^{-}\right]=\frac{K_{\mathrm{W}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}$

$\left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.4\times {10}^{-9}\mathrm{M}}$

$\left[{\mathrm{OH}}^{-}\right]=7.1\times {10}^{-6}\mathrm{M}$

Step 8: Explanation part(b)

Find the result of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ and $\left[{\mathrm{OH}}^{-}\right]$, the value of $\left[{\mathrm{OH}}^{-}\right]$ in this case  the solution is simple.

$7.1\times {10}^{-6}>1.4\times {10}^{-9}$

$\left[{\mathrm{OH}}^{-}\right]>\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\to Basic$

The concentration of $\left[{\mathrm{OH}}^{-}\right]$ in the basic solution is greater .

As a result, the formula is basic.

Step 9: Given information part(c)

To find whether the given solutions is acidic, basic, or neutral.

Step 10: Explanation part(c)

Because of this, the concentration of $\left[{\mathrm{OH}}^{-}\right]$as $3.5\times {10}^{-10}\mathrm{M}$.

Find the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ by substituting the values of ionic product of water and concentration of ${\mathrm{OH}}^{-}$ in the equations of ionic product of water.

$K_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{K_{\mathrm{W}}}{\left[{\mathrm{OH}}^{-}\right]}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1.0\times {10}^{-14}}{8.0\times {10}^{-3}\mathrm{M}}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=1.3\times {10}^{-12}\mathrm{M}$

Step 11: Explanation part(c)

Check the value of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ and $\left[{\mathrm{OH}}^{-}\right]$ ,the value is $\left[{\mathrm{OH}}^{-}\right]$ such that the solution is basic.

$8.0\times {10}^{-3}>1.3\times {10}^{12}$

$\left[{\mathrm{OH}}^{-}\right]>\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\to Basic$

The concentration of $\left[{\mathrm{OH}}^{-}\right]$ in the basic solution is maximum.

As a conclusion, the solution is basic.

Step 12: Given information part(d)

To find whether the given solutions is acidic, basic, or neutral.

Step 13: Explanation part(d)

For this, the concentration of $\left[{\mathrm{OH}}^{-}\right]$as $3.5\times {10}^{-10}\mathrm{M}$.

Measure the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ by substituting the values of ionic product of water and concentration of ${\mathrm{OH}}^{-}$ in the formula of ionic product of water.

$K_{\mathrm{W}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{K_{\mathrm{W}}}{\left[{\mathrm{OH}}^{-}\right]}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{1.0\times {10}^{-14}}{3.5\times {10}^{-10}\mathrm{M}}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.0\times {10}^{-5}\mathrm{M}$

Step 14: Explanation part(d)

Find the values of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$and $\left[{\mathrm{OH}}^{-}\right]$ , in this case, the value is $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$, this indicates that the solution is acidic.

$3.0\times {10}^{-5}>3.5\times {10}^{-10}$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]>\left[{\mathrm{OH}}^{-}\right]\to Basic$

The quantity of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ in acidic solution is higher.

As a result, the solution has an acidic $pH$.