Strong acid-strong base titration

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The simplest form of a titration to work with is one where a strong acid is added to a strong base or vice versa. Since both strong acids and strong bases ionize completely in water, the basic reaction that occurs is a simple neutralization

H⁺(aq) + OH^-(aq) -> H₂O(l)

To figure out the pH of the solution at any point during a strong acid-strong base titration, simply work out how much H⁺ and OH^- have been added to the solution, do stoichiometry to figure out which is in excess, then compute the pH/pOH from that. (If the amounts of H⁺ and OH^- are equal, the titration is at its equivalence point and the pH is 7.0.)

Example: You titrate a solution of 100 ml of 0.100 M Ca(OH)², a strong base with 0.100 M HCl, a strong acid. If you add 75 ml of acid, what is the pH of the solution?

Solution: Since Ca(OH)₂ is a strong base, it is fully ionized

Ca(OH)₂ -> Ca⁺² + 2OH^-

We start with 0.100 L * 0.100 M = 0.0100 moles of Ca(OH)₂, which according to the above equation generates 0.0200 moles of OH^- ion

We add 0.075L * 0.100 M = 0.0075 moles of H⁺. This will form 0.0075 moles of H⁺ which then reacts with the 0.0200 moles of OH^-.

This is a limiting reagent problem, but a simple one. By inspection, it is easy to see that the H⁺ will run out first. The 0.0075 moles of H⁺ will react with 0.0075 moles of OH^-, leaving 0.0125 moles of OH^- ion behind. We started will 100ml of solution and added 75 ml, so we now have 175ml solution, and the [OH^-] is

[OH^-] = 0.0125 mol/0.175L = 0.0714 M

From this, we can get the pOH and thus the pH

pOH = -log₁₀([OH^-])

pOH = -log₁₀(0.0714) = 1.15

pH = 14-pOH = 12.85

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