Chemistry / Unit 10: Solutions, Acids and Bases
10.6 Neutralization and Titration
Notes
In neutralization, an acid and a base react to form a salt and water: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$.
- A titration finds an unknown concentration by adding a solution of known concentration (the titrant) until the reaction is complete.
- The equivalence point is when moles of $\text{H}^+$ equal moles of $\text{OH}^-$. An indicator such as phenolphthalein changes color near it (the endpoint).
- For 1 : 1 acids and bases: $M_aV_a = M_bV_b$.
- Otherwise, use moles and the mole ratio. $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ needs 2 mol NaOH per mol acid.
- Example: 25.0 mL HCl needs 20.0 mL of 0.150 M NaOH. $\displaystyle M_a = \frac{(0.150)(20.0)}{25.0} = 0.120$ M.
Practice
What are the products of $\text{HNO}_3 + \text{KOH}$?
Neutralization gives a salt ($\text{K}^+$ with $\text{NO}_3^-$) and water.
30.0 mL of 0.200 M NaOH exactly neutralizes 10.0 mL of $\text{H}_2\text{SO}_4$. What is the molarity of the $\text{H}_2\text{SO}_4$?
mol NaOH $= 0.200 \times 0.0300 = 0.00600$. mol $\displaystyle \text{H}_2\text{SO}_4 = \frac{0.00600}{2} = 0.00300$. $\displaystyle M = \frac{0.00300}{0.0100} = 0.300$ M. 0.600 M ignores the 1 : 2 ratio.
15.0 mL of 0.200 M HCl is titrated with 0.100 M NaOH. What volume of NaOH reaches the equivalence point?
$\displaystyle V_b = \frac{(0.200)(15.0)}{0.100} = 30.0$ mL.
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