Chemistry · General chemistry II · Worked example
Find the pH during a strong acid–base titration
25.0 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Find the pH at the start, after 10.0 mL of base, at the equivalence point and after 35.0 mL.
At the start
HCl is a strong acid, so [H₃O⁺] = 0.100 M and pH = 1.00. The flask holds 2.50 × 10⁻³ mol of acid, so equivalence comes at 25.0 mL of base.
Before equivalence
10.0 mL of base is 1.00 × 10⁻³ mol of OH⁻, which neutralizes that much acid. The 1.50 × 10⁻³ mol left is now in 35.0 mL, so [H₃O⁺] = 0.0429 M and pH = 1.37.
At equivalence
After 25.0 mL all the acid has reacted. Only water and NaCl remain, and neither affects the pH, so pH = 7.00 at 25 °C.
Past equivalence
After 35.0 mL, 1.00 × 10⁻³ mol of OH⁻ is left over in 60.0 mL: [OH⁻] = 0.0167 M, pOH = 1.78 and pH = 12.22.
Read the curve
The pH rises only from 1.00 to 1.37 over the first 10.0 mL, then leaps through 7.00 within a drop or two of 25.0 mL and levels off in the basic range.
| NaOH added (mL) | 0 | 10.0 | 25.0 | 35.0 |
|---|---|---|---|---|
| pH | 1.00 | 1.37 | 7.00 | 12.22 |
Result
pH 1.00 at the start, 1.37 after 10.0 mL, 7.00 at equivalence (25.0 mL) and 12.22 after 35.0 mL.
Your turn
Find the pH after 20.0 mL of base has been added.
Show the answer and explanation
pH = 1.95.
2.50 × 10⁻³ − 2.00 × 10⁻³ = 5.0 × 10⁻⁴ mol of H₃O⁺ remains in 45.0 mL, so [H₃O⁺] = 0.011 M and pH = 1.95.
Keep exploring
Buffers & titration curves opens after 10.0 mL of base, at pH 1.37. Set the added volume to 35.0 mL to read 12.22 past equivalence.
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