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Calculating pH of Weak Acid and Base Solutions

Calculations used to determine pH in solutions of weak acids or bases

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Calculating pH of Weak Acid and Base Solutions

Credit: Jon Sullivan
Source: http://commons.wikimedia.org/wiki/File:Bees_pollenating_basil.jpg

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Calculating pH of Weak Acid and Base Solutions

The Ka\begin{align*}K_a\end{align*} and Kb\begin{align*}K_b\end{align*} values have been determined for a great many acids and bases, as shown in Tables 21.5 and 21.6. These can be used to calculate the pH of any solution of a weak acid or base whose ionization constant is known.

Sample Problem: Calculating the pH of a Weak Acid

Calculate the pH of a 2.00 M solution of nitrous acid (HNO2). The Ka\begin{align*}K_a\end{align*} for nitrous acid is 4.5 × 10-4.

Step 1: List the known values and plan the problem.

Known

• initial [HNO2] = 2.00 M
• Ka=4.5×104\begin{align*}K_a=4.5 \times 10^{-4}\end{align*}

Unknown

• pH = ?

First, an ICE table is set up with the variable x\begin{align*}x\end{align*} used to signify the change in concentration of the substance due to ionization of the acid. Then the Ka\begin{align*}K_a\end{align*} expression is used to solve for x\begin{align*}x\end{align*} and calculate the pH.

Step 2: Solve.

 Concentrations [HNO2] [H+] [NO2−] Initial 2.00 0 0 Change −x\begin{align*}-x\end{align*} +x\begin{align*}+x\end{align*} +x\begin{align*}+x\end{align*} Equilibrium 2.00−x\begin{align*}2.00-x\end{align*} x\begin{align*}x\end{align*} x\begin{align*}x\end{align*}

The Ka\begin{align*}K_a\end{align*} expression and value is used to set up an equation to solve for x\begin{align*}x\end{align*}.

Ka=4.5×104=(x)(x)2.00x=x22.00x\begin{align*}K_a=4.5 \times 10^{-4}=\frac{(x)(x)}{2.00 - x}=\frac{x^2}{2.00 - x}\end{align*}

The quadratic equation is required to solve this equation for x\begin{align*}x\end{align*}. However, a simplification can be made because of the fact that the extent of ionization of weak acids is small. The value of x\begin{align*}x\end{align*} will be significantly less than 2.00, so the "x\begin{align*}-x\end{align*}” in the denominator can be dropped.

4.5×104x=x22.00xx22.00=4.5×104(2.00)=2.9×102 M=[H+]\begin{align*}4.5 \times 10^{-4}&=\frac{x^2}{2.00 - x} \approx \frac{x^2}{2.00} \\ x&=\sqrt{4.5 \times 10^{-4}(2.00)}=2.9 \times 10^{-2} \ \text{M}= \left [ H^+ \right ]\end{align*}

Since the variable x\begin{align*}x\end{align*} represents the hydrogen-ion concentration, the pH of the solution can now be calculated.

pH=log[H+]=log[2.9×102]=1.54\begin{align*}pH=- \log[\text{H}^+]=- \log[2.9 \times 10^{-2}]=1.54\end{align*}

The pH of a 2.00 M solution of a strong acid would be equal to log(2.00)=0.30\begin{align*}- \log(2.00) = -0.30\end{align*}. The higher pH of the 2.00 M nitrous acid is consistent with it being a weak acid and therefore not as acidic as a strong acid would be.

The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the sample problem. However, the variable x\begin{align*}x\end{align*} will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.

Review

1. What does x\begin{align*}x\end{align*} stand for in the equation?
2. What simplifying assumption is made?
3. What would x\begin{align*}x\end{align*} stand for if we were calculating pOH?

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