Question

What is the H+ concentration for an aqueous solution with pOH = 3.58 at 25 degree C? Express your answer to two significant figures and include the appropriate units. Arrange the following aqueous solutions, all at 25 degree C, in order of decreasing acidity. Rank from most acidic to most basic. To rank items as equivalent, overlap them. At a certain temperature, the pH of a neutral solution is 7.21. What is the value of Kw, at that temperature? Express your answer numerically using two significant figures.

Answer 1

Concepts and reason

The concept used to solve this problem is based on pH of a solution.

Any substance that increase the concentration of ${{\rm{H}}^ + }$ is defined as an acid while a base increase concentration of ${\rm{O}}{{\rm{H}}^ - }$ . The concentrations of these ions can vary over a wide range. To avoid dealing these numbers, the concentrations of ${{\rm{H}}^ + }$ and ${\rm{O}}{{\rm{H}}^ - }$ are reported in terms of pH and pOH.

Fundamentals

The pH of a solution is calculated as follow.

${\rm{pH}} = - \log \left[ {{{\rm{H}}^ + }} \right]$

Here, $\left[ {{{\rm{H}}^ + }} \right]$ is the concentration of ${{\rm{H}}^ + }$ .

And, pOH is calculated as follow.

${\rm{pH}} = - \log \left[ {{\rm{O}}{{\rm{H}}^ - }} \right]$

Here, $\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]$ is concentration of ${\rm{O}}{{\rm{H}}^ - }$ .

Part 1

The relation between pH and pOH is as follows:

${\rm{pH}} + {\rm{pOH}} = 14$

Rearrange this equation and substitute 3.58 for pOH.

$\begin{array}{c}\\{\rm{pH}} = 14 - {\rm{pOH}}\\\\{\rm{ = 14}} - {\rm{3}}{\rm{.58}}\\\\{\rm{ = 10}}{\rm{.42}}\\\end{array}$

Substitute 10.42 in pH equation.

$\begin{array}{c}\\{\rm{pH}} = - \log \left[ {{{\rm{H}}^ + }} \right]\\\\10.42 = - \log \left[ {{{\rm{H}}^ + }} \right]\\\\\left[ {{{\rm{H}}^ + }} \right] = {\rm{antilog}}\left( { - 10.42} \right)\\\\\left[ {{{\rm{H}}^ + }} \right] = 38.02 \times {10^{ - 12}}{\rm{ M}}\\\end{array}$

Part 2

Calculate the pH from the concentration of HCl in pH equation. Thus, substitute 0.0023 M for ${{\rm{H}}^ + }$ .

$\begin{array}{c}\\{\rm{pH}} = - \log \left[ {0.0023{\rm{ M}}} \right]\\\\{\rm{ }} = - \left( { - {\rm{2}}{\rm{.64}}} \right)\\\\{\rm{ }} = {\rm{2}}{\rm{.64}}\\\end{array}$

Now, the relation between pH and the pOH is as follows:

${\rm{pH}} + {\rm{pOH}} = 14$

Thus, for third solution rearrange this equation and substitute 8.55 for pOH.

$\begin{array}{c}\\{\rm{pH}} = 14 - {\rm{pOH}}\\\\{\rm{ = 14}} - 8.55\\\\{\rm{ = 5}}{\rm{.45}}\\\end{array}$

Now, for fourth solution substitute 0.0018 M for ${\rm{O}}{{\rm{H}}^ - }$ in pOH equation.

$\begin{array}{c}\\{\rm{pOH}} = - \log \left[ {0.0018{\rm{ M}}} \right]\\\\{\rm{ }} = - \left( { - {\rm{2}}{\rm{.75}}} \right)\\\\{\rm{ }} = {\rm{2}}{\rm{.75}}\\\end{array}$

Now, the relation between pH and the pOH is as follows:

${\rm{pH}} + {\rm{pOH}} = 14$

Rearrange this equation and substitute 2.75 for pOH.

$\begin{array}{c}\\{\rm{pH}} = 14 - {\rm{pOH}}\\\\{\rm{ }} = {\rm{14}} - 2.{\rm{75}}\\\\{\rm{ }} = {\rm{11}}{\rm{.25}}\\\end{array}$

Now, compare the four values of pH. Thus, the order of four solutions in decreasing acidity is as follows:

$1 > 2 = 3 > 4$

Part 3

First convert the pH into ${{\rm{H}}^ + }$ concentration. Then, substitute 7.21 for pH.

$\begin{array}{c}\\{\rm{pH}} = - \log \left[ {{{\rm{H}}^ + }} \right]\\\\7.21 = - \log \left[ {{{\rm{H}}^ + }} \right]\\\\\left[ {{{\rm{H}}^ + }} \right] = {\rm{antilog}}\left( { - 7.21} \right)\\\\\left[ {{{\rm{H}}^ + }} \right] = 61.80 \times {10^{ - 9}}{\rm{ M}}\\\end{array}$

Since, the solution is neutral thus,

${\rm{pH}} = {\rm{pOH}}$

Therefore, the ionic product of water is written as follows:

$\begin{array}{c}\\{K_w} = \left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]\\\\ = {\left[ {{{\rm{H}}^ + }} \right]^2}\\\end{array}$

Substitute $61.80 \times {10^{ - 9}}{\rm{ M}}$ for $\left[ {{{\rm{H}}^ + }} \right]$ .

$\begin{array}{c}\\{K_w} = {\left[ {61.8 \times {{10}^{ - 9}}} \right]^2}\\\\{\rm{ = }}3.8 \times {10^{ - 15}}\\\end{array}$

Ans: Part 1

The concentration of ${{\rm{H}}^ + }$ is $38.02 \times {10^{ - 12}}{\rm{ M}}$ .

Part 2

The order of four solutions in decreasing acidity is as follows:

$1 > 2 = 3 > 4$

Part 3

The ionic product of water is equal to $3.8 \times {10^{ - 15}}$ .