14.1.1 pH and pOH

From ChemEd Collaborative

pH and pOH

The calculations we have just done show that the concentrations of hydronium and hydroxide ions in aqueous solution can vary from about 1 mol dm^–3 down to about 1 × 10^–14 mol dm^–3, and perhaps over an even wider range. The numbers used to express [H₃O⁺] and [OH^–] in the units mole per cubic decimeter will often include large negative powers of 10. Consequently it is convenient to define the following:

$\begin{align} & \text{pH}=-\text{log}\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{1 mol dm}^{-\text{3}}}\text{ } \\ & \\ & \text{pOH}=-\text{log}\frac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{\text{1 mol dm}^{-\text{3}}}\text{ } \\ \end{align}$

Note carefully what these equations tell us to do. To obtain pH, for example, we divide [H₃O⁺] by the units mole per cubic decimeter. 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 [H₃O⁺] and [OH^–] are usually between 1 and 10^-14 pH and pOH values are usually between 0 and 14.

EXAMPLE 14.2 Calculate the pH and the pOH of each of the following aqueous solutions: (a) 1.00 M HNO₃; 0.306 M Ba(OH)₂ .

Solution

a) Our previous discussion showed that for this solution [H₃O⁺] = 1.00 mol dm^–3 and [OH^–] = 1.00 . Applying the definitions of pH and pOH, we have

$\text{pH}=-\text{log}\frac{\text{1}\text{.00 mol dm}^{-\text{3}}}{\text{1 mol dm}^{-\text{3}}}=-\text{log(10 }^{\text{0}}\text{)}=-\text{(0)}=\text{0}\text{.00}$

$\text{pOH}=-\text{log}\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol dm}^{-\text{3}}}{\text{1 mol dm}^{-\text{3}}}=-\text{log(10}^{-\text{14}}\text{)}=-\text{(}-\text{14)}=14.\text{00}$

b) In Example 14.1 we found for this solution [H₃O⁺] = 1.63 × 10^–14 mol dm^–3 and [OH^–] = 6.12 × 10^–1 mol dm^–3. Thus

pH = –log(1.63 × 10^–14) = –[log(1.63) + log(10^–14)] = –[0.212 +(–14)] = –(–13.788) = 13.788

pOH = –log(6.12 × 10^–1) = –[log(6.12) + log(10^–1)] = –[0.787 + (–1)] = –(– 0.213) = 0.213

Note that to obtain the logarithm of a number which is not an exact power of 10, we first express the number in scientific notation. The logarithm of the power of 10 is just the exponent; that is, log(10^–14) = –14, and the logarithm of the number multiplied times the power of 10 can be obtained from a table of logarithms. If you have a calculator with a log key, all these steps are not necessary―simply enter the number (after dividing out the units) and hit the log key. Do not forget to change the sign, though, since it is minus the logarithm that you want.

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. Suffice it to say that unless 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 14.2b 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 [H₃O⁺]. This is the reverse of finding pH from [H₃O⁺]. Consequently it involves antilogs instead of logs. From the definition

$\text{pH}=-\text{log}\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}$

we have

$-\text{pH}=\text{log}\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}$

Taking the antilog of both sides, we have

$\text{antilog}\left( -\text{pH} \right)\text{ = antilog}-\text{pH}=\left\{ \text{log}\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}} \right\}$

so that $\text{antilog}\left( -\text{pH} \right)=\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}$

remembering that antilog x = 10^x, we can write this expression as

$\text{10}^{-\text{pH}}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}$

or [H₃O⁺] = 10^–pH mol dm^–3 (14.2a)

An alternative method of writing this equation is

$\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }=\frac{\text{1}}{\text{10}^{\text{pH}}}\text{ mol dm}^{-\text{3}}$ (14.2b)

EXAMPLE 14.3 The pH of a solution is found to be 3.40. Find the hydronium-ion 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 – 04 is the number of moles of hydronium ion per cubic decimeter. This follows from Eq. (14.2a):

[H₃O⁺] = 10^–pH mol dm^–3 = 10^–3.4 mol dm^–3 = 3.98 × 10^–4 mol dm^–3

The same result is almost as easy to find using Eq. (14.2b) and a table of logarithms.

10^–pH = antilog (pH) = antilog 3.40 = antilog 3 × antilog 0.40 = 10³ × 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,

[H₃O⁺] = 3.98 10^–4 mol dm^–3

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 (55.5 mol dm^–3)² [H₃O⁺][OH^–] = K_w = 10^–14 mol² dm^–6

Dividing both sides by mol² dm^–6, we obtain

$\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}\text{ }\times \text{ }\frac{[\text{OH}^{-}]}{\text{mol dm}^{-\text{3}}}=\text{10}^{-\text{14}}$

Taking logs and multiplying both by – 1, we then have

$-\text{log}\frac{\text{ }\!\![\!\!\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }\!\!]\!\!\text{ }}{\text{mol dm}^{-\text{3}}}\text{ }-\text{ }\log \text{ }\frac{[\text{OH}^{-}]}{\text{mol dm}^{-\text{3}}}=-\text{log}(\text{10}^{-\text{14}}\text{)}$

or pH + pOH = 14.00 (14.3)

This simple relationship is often useful in finding the pH of solutions containing bases, as the following example shows.

EXAMPLE 14.4 If 3.53 g of pure NaOH is dissolved in 10 dm³ of H₂O 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 $c_{\text{NaOH}}=\frac{n_{\text{NaOH}}}{V}=\frac{\text{0}\text{.088 25 mol}}{\text{10 dm}^{\text{3}}}=\text{8}\text{.82 }\times \text{ 10}^{\text{3}}\text{ mol dm}^{-\text{3}}\text{ }$

Since NaOH is a strong base, each mole of NaOH dissolved produces 1 mol OH^– ions, so that

[OH^–] = 8.82 × 10^–3 mol dm^–3

Thus pOH = –log(8.82 × 10^–3) = – (0.95 – 3.00) = +2.05

From which pH = 14.00 – pOH = 11.95

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14.1.1 pH and pOH

From ChemEd Collaborative

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