**Ouch, that hurts!**

Bees are beautiful creatures that help plants flourish. They carry pollen from one plant to another to facilitate plant growth and development. But, they can also be troublesome when they sting you. For people who are allergic to bee venom, this can be a serious, life-threatening problem. For the rest of us, it can be a painful experience. When stung by a bee, one first-aid treatment is to apply a paste of baking soda (sodium bicarbonate) to the stung area. This weak base helps with the itching and swelling that accompanies the bee sting.

### Calculating pH of Weak Acid and Base Solutions

The \begin{align*}K_a\end{align*}

#### Sample Problem: Calculating the pH of a Weak Acid

Calculate the pH of a 2.00 M solution of nitrous acid (HNO_{2}). The \begin{align*}K_a\end{align*}^{-4}.

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

Known

- initial [HNO
_{2}] = 2.00 M - \begin{align*}K_a=4.5 \times 10^{-4}\end{align*}
Ka=4.5×10−4

Unknown

- pH = ?

First, an ICE table is set up with the variable \begin{align*}x\end{align*}

*Step 2: Solve.*

Concentrations |
[HNO_{2}] |
[H^{+}] |
[NO_{2}^{−}] |

Initial | 2.00 | 0 | 0 |

Change | \begin{align*}-x\end{align*} |
\begin{align*}+x\end{align*} |
\begin{align*}+x\end{align*} |

Equilibrium | \begin{align*}2.00-x\end{align*} |
\begin{align*}x\end{align*} |
\begin{align*}x\end{align*} |

The \begin{align*}K_a\end{align*}

\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 \begin{align*}x\end{align*}

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

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

*Step 3: Think about your result.*

The pH of a 2.00 M solution of a strong acid would be equal to \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 \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

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