Q. 30 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. $\left[H_3{O}^{+}\right]=6.0\times {10}^{-12}M$

b. $\left[H_3{O}^{+}\right]=1.4\times {10}^{-4}M$

c. $\left[O{H}^{-}\right]=5.0\times {10}^{-12}M$

d. $\left[O{H}^{-}\right]=4.5\times {10}^{-2}M$

Step-by-Step Solution

Answer

a. The mixture is basic.

b. The mixture is acidic.

c. The mixture is acidic.

d. The mixture is basic.

Step 1: Introduction (part a)

The given is $\left[H_3{O}^{+}\right]=6.0\times {10}^{-12}M$.

We have to find whether the solution is acidic, basic or neutral.

Step 2: Explanation (part a)

Fill in the $K_w$ for water.

$K_w=\left[H_3{O}^{+}\right]\times \left[O{H}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the $K_w$ for the best results

$\frac{K_w}{\left[H_3{O}^{+}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]\times \left[O{H}^{-}\right]}{\cancel{[H_3{O}^{+}}]}=\frac{1.0\times {10}^{-14}}{\left[H_3{O}^{+}\right]}$

$\left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}}{\left[H_3{O}^{+}\right]}$

Calculate by substituting the known $\left[H_3{O}^{+}\right]$

$\left[O{H}^{-}\right]=\frac{2.0\times {10}^{-14}}{[6.0\times {10}^{-12}]}=2.0\times {10}^{-3}M$

The mixture is basic because the $\left[O{H}^{-}\right]$ of $2.0\times {10}^{-3}M$ is greater than the $\left[H_3{O}^{+}\right]$ of $6.0\times {10}^{-12}M$.

Step 3: Introduction (part b)

The given is $\left[H_3{O}^{+}\right]=1.4\times {10}^{-4}M$.

We have to find whether the solution is acidic, basic or neutral.

Step 4: Explanation (part b)

Fill in the $K_w$ for water.

$K_w=\left[H_3{O}^{+}\right]\times \left[O{H}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the $K_w$ for the best results.

$\frac{K_w}{\left[H_3{O}^{+}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]\times \left[O{H}^{-}\right]}{\cancel{[H_3{O}^{+}}]}=\frac{1.0\times {10}^{-14}}{\left[H_3{O}^{+}\right]}$

$\left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}}{\left[H_3{O}^{+}\right]}$

Calculate by substituting the known $\left[H_3{O}^{+}\right]$

$\left[O{H}^{-}\right]=\frac{1.0\times {10}^{-14}}{[1.4\times {10}^{-4}]}=7.0\times {10}^{-11}M$

The mixture is acidic because the $\left[H_3{O}^{+}\right]$ of $1.4\times {10}^{-4}M$ is greater than the

$\left[O{H}^{-}\right]$ of $7.0\times {10}^{-11}M$.

Step 5: Introduction (part c)

The given is $\left[O{H}^{-}\right]=5.0\times {10}^{-12}M$.

We have to find whether the solution is acidic, basic or neutral.

Step 6: Explanation (part c)

Fill in the $K_w$ for water.

$K_w=\left[H_3{O}^{+}\right]\times \left[O{H}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the $K_w$ for the best results.

$\frac{K_w}{\left[O{H}^{-}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]\times \left[O{H}^{-}\right]}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{\left[O{H}^{-}\right]}$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{\left[O{H}^{-}\right]}$

Calculate by substituting the known $\left[O{H}^{-}\right]$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{[5.0\times {10}^{-12}]}=2.0\times {10}^{-3}M$

The mixture is acidic because the $\left[H_3{O}^{+}\right]$ of $2.0\times {10}^{-3}$ is greater than $\left[O{H}^{-}\right]$ of $5.0\times {10}^{-12}M$

Step 7: Introduction (part d)

The given is $\left[O{H}^{-}\right]=4.5\times {10}^{-2}M$.

We have to find whether the solution is acidic, basic or neutral.

Step 8: explanation (part d)

Fill in the $K_w$ for water.

$K_w=\left[H_3{O}^{+}\right]\times \left[O{H}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the $K_w$ for the best results.

$\frac{K_w}{\left[O{H}^{-}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]\times \left[O{H}^{-}\right]}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{\left[O{H}^{-}\right]}$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{\left[O{H}^{-}\right]}$

Calculate by substituting the known $\left[O{H}^{-}\right]$

$\left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{[4.0\times {10}^{-2}]}=2.0\times {10}^{-13}M$

The mixture is basic because the $\left[O{H}^{-}\right]$ of $4.5\times {10}^{-2}$ is greater than $\left[H_3{O}^{+}\right]$ of $2.0\times {10}^{-13}M$.