91Ó°ÊÓ

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

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

Fill in the $K_w$for water.

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

Rearrange the $K_w$for the best results

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

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

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

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

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

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

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

Fill in the $K_w$for water.

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

Rearrange the $K_w$for the best results.

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

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

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

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

The mixture is acidic because the localid="1652591787750" $\left[H_3{O}^{+}\right]$of localid="1652591806453" $1.4×{10}^{-4}M$is greater than the

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

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

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

Fill in the $K_w$for water.

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

Rearrange the $K_w$for the best results.

localid="1652592906507" $\frac{K_w}{\left[O{H}^{-}\right]}=\frac{\left[\cancel{H_3{O}^{+}}\right]×\left[O{H}^{-}\right]}{\left[O{H}^{-}\right]}=\frac{1.0×{10}^{-14}}{\left[O{H}^{-}\right]}$

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

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

localid="1652592462495" $\left[H_3{O}^{+}\right]=\frac{1.0×{10}^{-14}}{[5.0×{10}^{-12}]}=2.0×{10}^{-3}M$

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

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

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

Fill in the $K_w$for water.

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

Rearrange the $K_w$for the best results.

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

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

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

role="math" localid="1652593152966" $\left[H_3{O}^{+}\right]=\frac{1.0×{10}^{-14}}{[4.0×{10}^{-2}]}=2.0×{10}^{-13}M$

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