Chemistry: Titration Data and Endpoint Analysis
Using titration data to find concentration, equivalence point, and endpoint accuracy
Chemistry: Titration Data and Endpoint Analysis
Using titration data to find concentration, equivalence point, and endpoint accuracy
Chemistry - Grade 9-12
- 1
A student titrates 25.00 mL of HCl with 0.1000 M NaOH. The endpoint is reached after 18.60 mL of NaOH is added. Write the balanced equation and calculate the molarity of the HCl.
For a strong acid and strong base with a 1:1 mole ratio, moles of acid equal moles of base at the equivalence point.
The balanced equation is HCl + NaOH -> NaCl + H2O. The moles of NaOH are 0.1000 M x 0.01860 L = 0.001860 mol. Because the mole ratio is 1:1, the HCl also has 0.001860 mol. The molarity of HCl is 0.001860 mol / 0.02500 L = 0.07440 M. - 2
A 20.00 mL sample of acetic acid, HC2H3O2, is titrated with 0.1500 M NaOH. The average volume of NaOH used is 16.40 mL. Calculate the molarity of the acetic acid. The reaction is HC2H3O2 + NaOH -> NaC2H3O2 + H2O.
The moles of NaOH are 0.1500 M x 0.01640 L = 0.002460 mol. The acid and base react in a 1:1 ratio, so the acetic acid has 0.002460 mol. The molarity of acetic acid is 0.002460 mol / 0.02000 L = 0.1230 M. - 3
A titration curve for a strong acid titrated with a strong base shows a very steep rise in pH near 24.8 mL of base added. Identify the approximate equivalence point volume and explain your choice.
Look for the middle of the sharpest pH change, not simply the first visible color change.
The approximate equivalence point volume is 24.8 mL. On a titration curve, the equivalence point occurs at the center of the steep vertical region where the pH changes most rapidly. - 4
A student records the following NaOH volumes for three titrations of the same acid sample: 22.41 mL, 22.38 mL, and 22.44 mL. Calculate the average volume of NaOH used.
The average volume is (22.41 mL + 22.38 mL + 22.44 mL) / 3 = 22.41 mL. The average NaOH volume used is 22.41 mL. - 5
In a titration, the indicator changes from colorless to faint pink. The pink color disappears after swirling. Has the endpoint been reached? Explain what the student should do next.
A temporary color change usually means the solution is close to the endpoint but not there yet.
The endpoint has not been reached because the pink color disappeared after swirling. The student should continue adding titrant drop by drop until a faint pink color remains for about 20 to 30 seconds. - 6
A 10.00 mL sample of H2SO4 is titrated with 0.2000 M NaOH. The endpoint requires 15.30 mL of NaOH. The balanced equation is H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Calculate the molarity of the H2SO4.
Use the coefficients in the balanced equation. Sulfuric acid reacts with twice as many moles of NaOH.
The moles of NaOH are 0.2000 M x 0.01530 L = 0.003060 mol. The mole ratio is 1 mol H2SO4 to 2 mol NaOH, so moles of H2SO4 are 0.003060 mol / 2 = 0.001530 mol. The molarity of H2SO4 is 0.001530 mol / 0.01000 L = 0.1530 M. - 7
A student accidentally overshoots the endpoint while titrating an acid with NaOH. The solution turns dark pink instead of faint pink. How will this affect the calculated acid concentration if the student uses the overshot volume? Explain.
The calculated acid concentration will be too high. Overshooting uses too much NaOH volume, which makes the calculated moles of NaOH too large. Since those moles are used to determine the moles of acid, the calculated acid concentration becomes too large. - 8
A buret reading is 1.25 mL at the start and 27.80 mL at the endpoint. What volume of titrant was delivered?
Buret readings increase as liquid leaves the buret.
The volume delivered is the final buret reading minus the initial buret reading. The volume delivered is 27.80 mL - 1.25 mL = 26.55 mL. - 9
A 25.00 mL sample of NaOH is titrated with 0.1250 M HCl. The endpoint is reached after 19.75 mL of HCl is added. Calculate the molarity of the NaOH.
The moles of HCl are 0.1250 M x 0.01975 L = 0.002469 mol. HCl and NaOH react in a 1:1 ratio, so the NaOH has 0.002469 mol. The molarity of NaOH is 0.002469 mol / 0.02500 L = 0.09875 M. - 10
A weak acid is titrated with a strong base. The equivalence point on the titration curve is above pH 7. Explain why the equivalence point is basic.
Think about what species remains in solution after all of the weak acid has reacted.
The equivalence point is basic because the weak acid has been converted into its conjugate base. The conjugate base reacts with water to produce hydroxide ions, which raises the pH above 7. - 11
A titration uses phenolphthalein, which changes color near pH 8.2 to 10.0. Is phenolphthalein a good indicator for a strong acid-strong base titration? Explain.
Phenolphthalein is a good indicator for a strong acid-strong base titration because the pH changes very rapidly near the equivalence point. Its color-change range falls within the steep part of the titration curve, so it gives an endpoint close to the equivalence point. - 12
The data table shows pH values during a titration of an unknown acid with NaOH: 0.00 mL, pH 2.10; 10.00 mL, pH 2.75; 20.00 mL, pH 3.80; 24.00 mL, pH 6.20; 25.00 mL, pH 8.70; 26.00 mL, pH 11.30; 30.00 mL, pH 12.10. Estimate the equivalence point volume and justify your answer.
Find the interval where the pH changes the most over the smallest added volume.
The equivalence point is about 25.00 mL. The largest and steepest pH change occurs between 24.00 mL and 26.00 mL, and the midpoint of that sharp change is approximately 25.00 mL. - 13
A student rinses a buret with water but does not rinse it with the NaOH solution before filling it for titration. How could this affect the calculated concentration of the acid being tested?
The calculated acid concentration could be too low. Water left in the buret dilutes the NaOH, so the actual NaOH concentration is lower than the value used in the calculation. Using the higher listed NaOH concentration can cause the calculated acid concentration to be inaccurate, usually too low if a larger volume is required but the assumed concentration is not corrected. - 14
A 25.00 mL sample of a monoprotic acid is titrated with 0.1100 M NaOH. The equivalence point occurs at 21.35 mL of NaOH. Calculate the moles of acid in the original sample and the acid molarity.
Monoprotic means each acid molecule donates one hydrogen ion, so the acid and NaOH mole ratio is 1:1.
The moles of NaOH are 0.1100 M x 0.02135 L = 0.002349 mol. Since the acid is monoprotic, it reacts with NaOH in a 1:1 ratio, so the original sample contained 0.002349 mol of acid. The acid molarity is 0.002349 mol / 0.02500 L = 0.09394 M. - 15
During a titration, a student reads the bottom of the meniscus at eye level for the initial reading but reads from above the meniscus for the final reading. Explain why this is a problem and how it affects data quality.
This is a problem because inconsistent viewing angles cause parallax error. Reading from above the meniscus can make the final buret reading appear different from its true value, which changes the calculated volume delivered. Since titration calculations depend directly on volume, the concentration result becomes less accurate.