14.3: pH and pOH
 Page ID
 49685
The calculations we have just done show that the concentrations of hydronium and hydroxide ions in aqueous solution can vary from about 1 mol L^{1}down to about 1 × 10^{–14} mol L^{–1}, and perhaps over an even wider range. The numbers used to express [H_{3}O^{+}] and [OH^{–}] in the units mole per liter will often include large negative powers of 10. Consequently it is convenient to define the following:
\[ \text{pH}=\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{1 mol L}^{\text{1}}}\\\text{ } \\ \text{pOH}=\text{log}\frac{\text{ }[\text{ OH}^{}\text{ }]\text{ }}{\text{1 mol L}^{\text{1}}}\text{ } \\ \]
Note carefully what these equations tell us to do. To obtain pH, for example, we divide [H_{3}O^{+}] by the units mole per liter. This gives a pure number, and so we can take its logarithm. (It does not make sense to take the logarithm of a unit, such as mole per cubic decimeter.) The minus sign insures that we will obtain a positive result most of the time.
The logarithm of a number is the power to which 10 must be raised to give the number itself. Therefore the definitions of pH and pOH mean that we can deal with powers of 10 rather than numerical values. Since the numbers needed to express [H_{3}O^{+}] and [OH^{–}] are usually between 1 and 10^{14} pH and pOH values are usually between 0 and 14.
Example \(\PageIndex{1}\): pH & pOH
Calculate the pH and the pOH of each of the following aqueous solutions: (a) 1.00 M HNO_{3}; 0.306 M Ba(OH)_{2} .
Solution
a) Our previous discussion showed that for this solution [H_{3}O^{+}] = 1.00 mol/L and [OH^{–}] = 1.00 x 10^{14}. Applying the definitions of pH and pOH, we have
\[\begin{align}\text{pH}=\text{log}\frac{\text{1}\text{.00 mol L}^{\text{1}}}{\text{1 mol L}^{\text{1}}}=\text{log(10 }^{\text{0}}\text{)}\\\text{ }=\text{(0)}=\text{0}\text{.00}\\\text{ }\\\text{pOH}=\text{log}\frac{\text{1}\text{.00 }\times \text{ 10}^{\text{14}}\text{ mol L}^{\text{1}}}{\text{1 mol L}^{\text{1}}}=\text{log(10}^{\text{14}}\text{)}\\\text{ }=\text{(}\text{14)}=14.\text{00}\end{align}\]
b)In the example in the section on ionization of water, we found for this solution [H_{3}O^{+}] = 1.63 × 10^{–14} mol/L and [OH^{–}] = 6.12 × 10^{–1} mol/L. Thus
\[\begin{align}\text{pH} = \text{log}{ 1.63}\times{10^{14}}=({13.788})=13.788\\\text{ }\\\text{pOH} = \text{log}{ 6.12}\times{10^{1}}=({0.213})=0.213\end{align}\]
In the laboratory it is convenient to measure the pH of a solution using a pH meter. Such a device works on a different principle from the conductivity measurements we have already mentioned, and an accurate explanation of how it works is beyond the scope of the present discussion. While greater accuracy can be obtained when great care and special instruments are used, pH is usually measured to an accuracy of ± 0.01. Therefore pH values are usually rounded to the second decimal place; the results of Example 1b would commonly be rounded to pH = 13.79 and pOH = 0.21.
Because pH measurements are so easily made, it is essential that you be able to convert from pH to [H_{3}O^{+}]. This is the reverse of finding pH from [H_{3}O^{+}]. Consequently it involves antilogs instead of logs. From the definition
\[\text{pH}=\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}}\]
we have
\[\text{pH}=\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}}\]
Taking the antilog of both sides, we have
\[\text{antilog}\left( \text{pH} \right)\text{ = antilog}\text{pH}=\left\{ \text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}} \right\}\]
so that
\[\text{antilog}\left( \text{pH} \right)=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}}\]
remembering that antilog x = 10^{x}, we can write this expression as
\[\text{10}^{\text{pH}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}}\]
or
\[\left[\text{H}_{3}\text{O}^{+}\right]=\text{10}^\text{pH}\text{ mol}\text{ L}^{1}\label{15}\]
An alternative method of writing this equation is
\[\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }=\frac{\text{1}}{\text{10}^{\text{pH}}}\text{ mol L}^{\text{1}}\]
Example \(\PageIndex{2}\): Hydronium Concentration
The pH of a solution is found to be 3.40. Find the hydroniumion concentration of the solution.
Solution
If you have a calculator which has an antilog or 10^{x} button, the problem is very simple. You enter – 3.40 and hit the button. The number thus obtained, 3.9822 x 10^{–4} is the number of moles of hydronium ion per liter. This follows from Eq. \(\ref{15}\):
\(\left[\text{H}_{3}\text{O}^{+}\right]=\text{10}^{\text{pH}}\text{ mol}\text{ L}^{1}=\text{10}^{3.4}\text{ mol}\text{ L}^{1}=\text{3.98}\times\text{10}^{4}\text{ mol}\text{ L}^{1}\)
The same result is almost as easy to find using Eq. (1b).
\(\text{10}^{\text{pH}}=\text{antilog}\left(\text{pH}\right)=\text{antilog 3.40}=\text{antilog 3}\times \text{antilog 0.40}=\text{10}^{3}\times \text{2.51}\)
Thus
\(\text{10}^{\text{pH}}=\frac{\text{1}}{\text{10}^{\text{pH}}}=\frac{\text{1}}{\text{2}\text{.51 }\times \text{ 10}^{\text{3}}}=\text{3}\text{.98 }\times \text{ 10}^{\text{4}}\)
in other words,
\(\left[\text{H}_{3}\text{O}^{+}\right]=\text{3.98}\times\text{10}^{4}\text{ mol}\text{ L}^{1}\)
There is a very simple relationship between the pH and the pOH of an aqueous solution at 25°C. We know that at this temperature
\[{K}_{w}={K}_{c}\left(\text{55.5 mol}\text{ L}^{1}\right)^{2}\left[\text{H}_{3}\text{O}^{+}\right] \left[\text{OH}^{}\right]={K}_{w}=\text{ 10}^{14} \text{mol}^{2}\text{L}^{2}\]
Dividing both sides by mol^{2} L^{–2}, we obtain
\[\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}}\text{ }\times \text{ }\frac{[\text{OH}^{}]}{\text{mol L}^{\text{1}}}=\text{10}^{\text{14}}\]
Taking logs and multiplying both by – 1, we then have
\[\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{\text{1}}}\text{ }\text{ }\log \text{ }\frac{[\text{OH}^{}]}{\text{mol L}^{\text{1}}}=\text{log}(\text{10}^{\text{14}}\text{)}\]
or
\[\text{pH} + \text{pOH}= \text{14.00}\]
This simple relationship is often useful in finding the pH of solutions containing bases, as the following example shows.
Example \(\PageIndex{3}\): pH of a Solution
If 3.53 g of pure NaOH is dissolved in 10 L of H_{2}O find the pH of the resulting solution.
Solution
We first calculate the concentration of the NaOH.
\(n_{\text{NaOH}}=\text{3}\text{.53 }\times \text{ }\frac{\text{1 mol}^{\text{1}}}{\text{40}\text{.0 g}}=\text{0}\text{.088 25 mol}\)
so that
\(c_{\text{NaOH}}=\frac{n_{\text{NaOH}}}{V}=\frac{\text{0}\text{.088 25 mol}}{\text{10 L}^{\text{1}}}=\text{8}\text{.82 }\times \text{ 10}^{\text{3}}\text{ mol L}^{\text{1}}\text{ }\)
Since NaOH is a strong base, each mole of NaOH dissolved produces 1 mol OH^{–} ions, so that
\(\left[\text{OH}^{}\right]=\text{8.82 }\times \text{10}^{3}\text{mol L}^{1}\)
Thus
\(\text{pOH}=\text{log }\left(\text{8.82}\times \text{10}^{3}\right)=\left(\text{0.95}\text{3.00}\right)=+\text{2.05}\)
From which
\(\text{pH} =\text{14.00  pOH} =\text{11.95}\)
While the ability to calculate the pH of a solution from the hydroniumion concentration and vice versa is useful, it is not the only thing we need to understand about pH. If someone gives you a solution whose pH is 14.74, it is true that the hydroniumion concentration must be 1.82 × 10^{–15} mol L^{–1} but it is perhaps more important to know that the solution is corrosively basic. In general, then, we need not only to be able to calculate a pH but also to have some realization of what kind of solutions have what kind of pH, as displayed in the following table. This table is part of our collection of acid and base constants.
Substance  pH  [H_{3}O^{+}]  [OH^{}]  pOH  Strength  

Battery acid 
0  1  10^{14}  14  Strongly acidic  


1  10^{1}  10^{13}  13  
2  10^{2}  10^{12}  12  
3  10^{3}  10^{11}  11  Weakly acidic  
Soda water  4  10^{4}  10^{10}  10  
Black coffee  5  10^{5}  10^{9}  9  Barely acidic  
6  10^{6}  10^{8}  8  
Pure water  7  10^{7}  10^{7}  7  Neutral  
Seawater  8  10^{8}  10^{6}  6  Barely basic  
Baking soda  9  10^{9}  10^{5}  5  
Toilette soap  10  10^{10}  10^{4}  4  Mildly basic  
Laundry water  11  10^{11}  10^{3}  3  
Household ammonia  12  10^{12}  10^{2}  2  Very basic  
13  10^{13}  10^{1}  1  
Drain cleaner  14  10^{14}  1  0 
In pure water at 25°C the hydroniumion concentration is close to 1.00 × 10^{–7} mol/L, so that the pH is 7. In consequence any solution, not only pure water, which has a pH of 7 is described as being neutral. An acidic solution, as we know, is one in which the hydroniumion concentration is greater than that of pure water, i.e., greater than 10^{–7} mol/L. In pH terms this translates into a pH which is less than 7 (because the pH is a negative logarithm). Small pH values are thus characteristic of acidic solutions; the smaller the pH, the more acidic the solution.
By contrast, a basic solution is one in which the hydroxideion concentration is greater than 10^{–7} mol/L. In such a solution the hydroniumion concentration is less than 10^{–7} mol/L, so that the pH of a basic solution is greater than 7. Large pH values are thus characteristic of basic solutions. The larger the pH, the more basic the solution.
Contributors
Ed Vitz (Kutztown University), John W. Moore (UWMadison), Justin Shorb (Hope College), Xavier PratResina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.