Let's **analyze your titration data step by step** to find the **volume of NaOH needed to reach the endpoint** (equivalence point) for the titration of hydrochloric acid (HCl) with sodium hydroxide (NaOH). --- ### **Step 1: Organize the Data** Your data is a bit jumbled, but let's try to reconstruct a table with the likely intended values: | mL NaOH Added | pH | |---------------|-------| | 0 | 2.4 | | 1 | 2.2 | | 2 | 2.3 | | 3 | 2.4 | | 4 | 2.5 | | 5 | 2.7 | | 6 | 3.2 | | 7 | 6.6 | | 8 | 10.8 | | 9 | 11.3 | | 10 | 11.3 | | 11 | 11.5 | | 12 | 11.6 | | 13 | 11.6 | *(I have interpolated some values for continuity, since your data had some repeats and formatting issues. Let me know if you need to adjust these.)* --- ### **Step 2: Identify the Endpoint (Equivalence Point)** - The **endpoint** is where the amount of NaOH added is stoichiometrically equal to the amount of HCl originally present. - On the **titration curve**, this is the point where the pH rises very sharply (steep vertical section). Let's look for this **sharp increase** in pH: - From 6 mL to 7 mL: pH jumps from **3.2 to 6.6** (large jump) - From 7 mL to 8 mL: pH jumps from **6.6 to 10.8** (very large jump) - After 8 mL, pH stays high, so we've passed the equivalence point. #### **Equivalence Point Estimate** - The **steepest increase** is between 7 mL and 8 mL. - Therefore, the equivalence point is **between 7 and 8 mL** of NaOH added. --- ### **Step 3: Refine the Estimate (Optional)** If you want a more precise value, you can average the two volumes, or interpolate: - **Average**: (7 mL + 8 mL) / 2 = **7.5 mL** But often, it's considered to be **at the point just before the sharp increase finishes**. So, **8 mL** is the best estimate with your data. --- ### **Step 4: Conclusion** **The volume of 0.02 M NaOH solution needed to reach the endpoint is approximately:** \[ \boxed{8.0 \ \text{mL}} \] --- ## **Summary Table** | mL NaOH Added | pH | |---------------|-------| | 6 | 3.2 | | 7 | 6.6 | ← Start of sharp jump | 8 | 10.8 | ← End of sharp jump (endpoint) | 9 | 11.3 | --- ### **Final Answer:** > **The endpoint (equivalence point) occurs at about 8.0 mL of 0.02 M NaOH added.** --- **If you have a graph, this would be the inflection point (vertical rise). If you need to calculate the amount of HCl present, let me know!**