Question

complete the tables of value for four aqueous solutions at 25 celcius

Answer 1

There are a total of four parameters you need to know to determine the pH of the solution. The pH and the pOH are negative logarithms of the concentrations of the ions H+ and OH-, respectively. These ions are the solvated ions of in an acid-base reaction. The equations are as follows:

pH = -log [H+]
pOH = -log [OH-]

Lastly, the pH range is from 1 to 14. So, it is easier to remember that pH + pOH is equal to 14. So, we use these three equations for the solutions.

Solution A:
pH = -log(6.4×10^-13) = 12.2
pOH = 14 - 12.2 = 1.8
1.8 = -log[OH-]
[OH-] = 0.016 M


Solution B:
pOH = -log(2.7×10^-10) = 9.6
pH = 14 - 9.6 = 4.4
4.4 = -log[H+]
[H+] = 3.98×10^-5 M


Solution C:
8.11 = -log(H+)
H+ = 7.76×10^-9
pOH = 14 - 8.11 = 5.89
5.89 = -log[OH-]
[OH-] = 1.288×10^-6 M


Solution D:
pH = 14 - 4.73 = 9.27
9.27= -log[H+]
[H+] = 5.37×10^-10 M
4.73 = -log[OH-]
[OH-] = 1.862×10^-5 M

Answer 2

The table of values for four aqueous solutions at   is as shown in Table 1.

Further Explanation:

The negative logarithm of hydrogen ion concentration is termed as pH while that of hydroxide ion concentration is pOH. Higher the pH of solution, more will be the alkalinity and vice-versa. Similarly, higher value of pOH indicates solution is acidic in nature .

The expression for pH is mentioned below.

$\text{pH}=-\text{log}\left[\text{H}^+\right]$                                             ...... (1)

Where $\left[\text{H}^+\right]$  is the concentration of hydrogen ion.

The expression for pOH is mentioned below.

$\text{pOH}=-\text{log}\left[\text{OH}^-\right]$                                        ...... (2)

Where $\left[\text{OH}^-\right]$  is the concentration of hydroxide ion.

pH and pOH are related to each other by following expression:

pH + pOH = 14                                               …… (3)

Solution A:

Substitute $6.4\times10^{-13}\text{ M}$  for $\left[\text{H}^+\right]$  in equation (1).

$\begin{aligned}\text{pH}=&-\text{log}\left[6.4\times10^{-13}\text{ M}\right]\\=&12.2\end{aligned}$

Rearrange equation (3) for pOH.

pOH = 14 – pH                                                                          …… (4)

Substitute 12.2 for pH in equation (4).

$\begin{aligned}\text{pOH}=&14-12.2\\=&1.8\end{aligned}$

Rearrange equation (2) for $\left[\text{OH}^-\right]$  .

$\left[\text{OH}^-\right]=10^{-\text{pOH}}$                                                                        …… (5)

Substitute 1.8 for pOH in equation (5).

$\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{1.8}}\\=&0.016\text{ M}\end{aligned}$

Solution B:

Substitute  $2.7\times10^{-10}\text{ M}$ for $\left[\text{OH}^-\right]$  in equation (2).

$\begin{aligned}\text{pOH}=&-\text{log}\left[2.7\times10^{-10}\text{ M}\right]\\=&\ 9.6\end{aligned}$

Rearrange equation (3) for pH.

pH = 14 – pOH                                                                                 …… (6)

Substitute 9.6 for pOH in equation (6).

$\begin{aligned}\text{pH}=&14-9.6\\=&4.4\end{aligned}$

Rearrange equation (1) for $\left[\text{H}^+\right]$ .

$\left[\text{H}^+\right]=10^\text{-pH}$                                                                                        …… (7)  

Substitute 4.4 for pH in equation (7).

$\begin{aligned}\left[\text{H}^+\right]=&10^\text{-4.4}\\=&0.0000398\text{ M}\end{aligned}$

Solution C:

Substitute 8.11 for pH in equation (7).

$\begin{aligned}\left[\text{H}^+\right]=&10^\text{-8.11}\\=&0.0000000078\text{ M}\end{aligned}$

Substitute 8.11 for pH in equation (4).

$\begin{aligned}\text{pOH}=&14-8.11\\=&5.89\end{aligned}$

Substitute 5.89 for pOH in equation (5).

$\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{5.89}}\\=&0.000001288\text{ M}\end{aligned}$

Solution D:

Substitute 4.73 for pOH in equation (5).

$\begin{aligned}\left[\text{OH}^-\right]=&10^{-\text{4.73}}\\=&0.00001862\text{ M}\end{aligned}$

Substitute 4.73 for pOH in equation (6).

$\begin{aligned}\text{pH}=&14-4.73\\=&9.27\end{aligned}$

Substitute 9.27 for pH in equation (7).

$\begin{aligned}\left[\text{H}^+\right]=&10^\text{-9.27}\\=&0.0000000005\text{ M}\end{aligned}$

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Answer details:

Grade: High School

Chapter: Acids, base and salts

Subject: Chemistry

Keywords: pH, pOH, H+, OH-, solution A, solution B, solution C, solution D, hydrogen ion, hydroxide ion, negative logarithm.