To solve pH and pOH questions without a log table, we typically work with concentrations that are exact powers of $10$. This simplifies the logarithm calculation. The key formulas are: • $\text{pH} = -\log[\text{H}^+]$ • $\text{pOH} = -\log[\text{OH}^-]$ • $\text{pH} + \text{pOH} = 14$ (at $25^\circ\text{C}$) • $[\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ (at $25^\circ\text{C}$) When the concentration of $\text{H}^+$ or $\text{OH}^-$ ions is in the form $1.0 \times 10^{-x}$ M, then $\log(1.0 \times 10^{-x}) = -x$. Example 1: Calculating pH from $[\text{H}^+]$ What is the pH of a solution where the hydrogen ion concentration, $[\text{H}^+]$, is $1.0 \times 10^{-4}$ M? Step 1: Use the pH formula. $$\text{pH} = -\log[\text{H}^+]$$ Step 2: Substitute the given concentration. $$\text{pH} = -\log(1.0 \times 10^{-4})$$ Step 3: Simplify the logarithm. Since $\log(10^{-4}) = -4$. $$\text{pH} = -(-4)$$ $$\text{pH} = 4$$ The pH of the solution is $\boxed{4}$. Example 2: Calculating pOH from $[\text{OH}^-]$ What is the pOH of a solution where the hydroxide ion concentration, $[\text{OH}^-]$, is $1.0 \times 10^{-2}$ M? Step 1: Use the pOH formula. $$\text{pOH} = -\log[\text{OH}^-]$$ Step 2: Substitute the given concentration. $$\text{pOH} = -\log(1.0 \times 10^{-2})$$ Step 3: Simplify the logarithm. Since $\log(10^{-2}) = -2$. $$\text{pOH} = -(-2)$$ $$\text{pOH} = 2$$ The pOH of the solution is $\boxed{2}$. Example 3: Calculating pH from pOH (or vice versa) If the pOH of a solution is $3$, what is its pH? Step 1: Use the relationship between pH and pOH. $$\text{pH} + \text{pOH} = 14$$ Step 2: Substitute the given pOH value. $$\text{pH} + 3 = 14$$ Step 3: Solve for pH. $$\text{pH} = 14 - 3$$ $$\text{pH} = 11$$ The pH of the solution is $\boxed{11}$.