Titration Curve Simulator

Every run titrates 25.0 mL of a 0.100 M analyte with 0.100 M titrant, so the equivalence point is always at 25.0 mL. The weak acid behaves like acetic acid ($K_a = 1.8 \times 10^{-5}$, $\text{p}K_a = 4.74$) and the weak base like ammonia ($K_b = 1.8 \times 10^{-5}$).

Start with Strong acid + strong base to see a curve that jumps through pH 7, then switch to Weak acid + strong base and compare where the curve starts, how flat the middle is, and where the jump lands.

The two base-in-flask options mirror the acid cases. With Weak base + strong acid, the curve starts high, falls through a buffer region where the triangle sits at pOH = pK_b (pH 9.26 for this base), and reaches an equivalence pH of 5.28, below 7, because the conjugate acid of the weak base is left in solution. With Strong base + strong acid, equivalence is again pH 7.00.

Work in moles first, then divide by the total volume. For a weak acid HA titrated with $\text{OH}^-$, the reaction $\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}$ goes to completion. What remains decides the method:

Titrate 25.0 mL of 0.100 M $\text{HC}_2\text{H}_3\text{O}_2$ ($K_a = 1.8 \times 10^{-5}$) with 0.100 M NaOH. Initial acid: $0.0250\ \text{L} \times 0.100\ \text{M} = 2.50$ mmol, so $V_{eq} = 25.0$ mL.

0 mL. $[\text{H}^+] = \sqrt{(1.8 \times 10^{-5})(0.100)} = 1.34 \times 10^{-3}$ M, pH = 2.87. ($x$ is 1.3% of 0.100, so the approximation holds.)

10.0 mL. 1.00 mmol $\text{OH}^-$ converts 1.00 mmol HA to $\text{A}^-$, leaving 1.50 mmol HA. $\text{pH} = 4.74 + \log\frac{1.00}{1.50} = 4.74 - 0.18 = 4.57$.

12.5 mL (half-equivalence). 1.25 mmol HA and 1.25 mmol $\text{A}^-$, so pH = pK_a = 4.74.

20.0 mL. 2.00 mmol $\text{A}^-$, 0.50 mmol HA: $\text{pH} = 4.74 + \log\frac{2.00}{0.50} = 4.74 + 0.60 = 5.35$.

25.0 mL (equivalence). 2.50 mmol $\text{A}^-$ in 50.0 mL gives 0.0500 M. $K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$; $[\text{OH}^-] = \sqrt{(5.6 \times 10^{-10})(0.0500)} = 5.3 \times 10^{-6}$ M; pOH = 5.28, pH = 8.72. The pH is above 7 because $\text{A}^-$ is a weak base.

30.0 mL. Excess $\text{OH}^-$ = 3.00 − 2.50 = 0.50 mmol in 55.0 mL = 0.00909 M; pOH = 2.04, pH = 11.96.

Set the menu to Weak acid + strong base and slide to each volume; the readout should match these values to within 0.01. Notice how little the pH moves between 10.0 mL and 20.0 mL (4.57 to 5.35, under one pH unit for 10 mL of base) compared with the jump of more than three units between 20.0 mL and 30.0 mL. That flat stretch is the buffer at work: added $\text{OH}^-$ is consumed by HA instead of staying free in solution.

Titration curves appear in Unit 8 questions that ask you to read pK_a from a graph, identify the major species at a given volume, explain why the buffer region resists pH change, or select an indicator. Expect to sketch or label a curve, mark the equivalence and half-equivalence points, and explain in words what particles are present at each one. A common setup gives a curve with no numbers for K_a and asks you to estimate it: read the pH at half the equivalence volume and take $K_a = 10^{-\text{pH}}$. The same reasoning in reverse gives K_b for a weak base titrated with strong acid.