Chemistry: Acid-Base Chemistry: Buffers and Titration Curves
Using buffer equations and interpreting acid-base titration graphs
Chemistry: Acid-Base Chemistry: Buffers and Titration Curves
Using buffer equations and interpreting acid-base titration graphs
Chemistry - Grade 9-12
- 1
A solution contains 0.20 M HF and 0.25 M NaF. Explain why this solution acts as a buffer.
A buffer usually contains a weak acid and its conjugate base, or a weak base and its conjugate acid.
This solution acts as a buffer because it contains a weak acid, HF, and its conjugate base, F-. The HF can neutralize added base, and the F- can neutralize added acid. - 2
A buffer contains 0.10 M acetic acid, HC2H3O2, and 0.25 M acetate ion, C2H3O2-. The pKa of acetic acid is 4.76. Calculate the pH of the buffer.
The acetate ion is the conjugate base, and acetic acid is the weak acid.
Using pH = pKa + log([base]/[acid]), pH = 4.76 + log(0.25/0.10). Since log(2.5) = 0.40, the pH is 5.16. - 3
A buffer is made with 0.200 mol NH3 and 0.300 mol NH4+ in 1.00 L of solution. The pKa of NH4+ is 9.25. What is the pH after 0.050 mol HCl is added? Assume the volume change is negligible.
Strong acid converts some of the weak base into its conjugate acid before using the Henderson-Hasselbalch equation.
HCl reacts with NH3, so NH3 decreases to 0.150 mol and NH4+ increases to 0.350 mol. Using pH = 9.25 + log(0.150/0.350), the pH is 9.25 + log(0.429), which equals about 8.88. - 4
A student says, "A buffer keeps pH from ever changing." Correct this statement using accurate chemistry language.
A buffer resists changes in pH, but it does not stop pH changes completely. If enough acid or base is added, the buffer can be overwhelmed and the pH will change significantly. - 5
A 25.0 mL sample of 0.100 M HCl is titrated with 0.125 M NaOH. Calculate the volume of NaOH needed to reach the equivalence point.
At the equivalence point for a monoprotic strong acid and strong base, moles of H+ equal moles of OH-.
The moles of HCl are 0.0250 L times 0.100 M, which equals 0.00250 mol. At the equivalence point, 0.00250 mol NaOH are needed. The volume of NaOH is 0.00250 mol divided by 0.125 M, which equals 0.0200 L, or 20.0 mL. - 6
A titration curve starts at pH 1.0, rises slowly, has a very steep vertical region centered at pH 7.0, and levels off above pH 12. What type of titration does this curve most likely show?
Look at both the initial pH and the pH at the equivalence point.
This curve most likely shows a strong acid being titrated with a strong base. The low starting pH shows a strong acid, and the equivalence point near pH 7 is typical for a strong acid and strong base titration. - 7
For a weak acid titrated with a strong base, the pH at the half-equivalence point is 4.74. What is the pKa of the acid, and what is its Ka?
At half-equivalence, the weak acid and conjugate base have equal concentrations.
At the half-equivalence point, pH equals pKa, so the pKa is 4.74. Ka equals 10^(-4.74), which is about 1.8 x 10^-5. - 8
A weak acid is titrated with NaOH. The equivalence point occurs at 32.0 mL of NaOH added. At what volume of NaOH added does the half-equivalence point occur? Explain what is special about that point.
The half-equivalence point occurs at 16.0 mL of NaOH added. At this point, half of the weak acid has been converted into its conjugate base, so the concentrations of acid and conjugate base are equal and pH equals pKa. - 9
Which indicator is more appropriate for a weak acid titrated with a strong base: methyl orange, which changes color around pH 3.1 to 4.4, or phenolphthalein, which changes color around pH 8.2 to 10.0? Explain your choice.
Choose an indicator whose color-change range falls within the steep vertical part of the titration curve.
Phenolphthalein is more appropriate because a weak acid titrated with a strong base has an equivalence point above pH 7. Its color-change range is closer to the steep region near the equivalence point. - 10
A 50.0 mL sample of 0.100 M acetic acid is titrated with 0.100 M NaOH. What volume of NaOH is needed to reach the equivalence point?
The moles of acetic acid are 0.0500 L times 0.100 M, which equals 0.00500 mol. Since acetic acid reacts with NaOH in a 1 to 1 ratio, 0.00500 mol NaOH are needed. The volume is 0.00500 mol divided by 0.100 M, which equals 0.0500 L, or 50.0 mL. - 11
The following titration data were collected for a weak acid titrated with 0.100 M NaOH: 0.0 mL, pH 2.9; 10.0 mL, pH 4.2; 20.0 mL, pH 4.8; 30.0 mL, pH 5.3; 40.0 mL, pH 8.7; 41.0 mL, pH 10.5; 50.0 mL, pH 11.5. Estimate the equivalence point volume and explain your reasoning.
The equivalence point is near the middle of the steepest vertical region of the curve.
The equivalence point is approximately 40.0 mL because the largest and steepest pH change occurs around that volume. The pH is above 7 at equivalence, which is consistent with a weak acid and strong base titration. - 12
A buffer contains equal concentrations of H2CO3 and HCO3-. The pKa for H2CO3 is 6.35. What is the pH of the buffer? Explain without doing a long calculation.
The pH is 6.35 because the acid and conjugate base concentrations are equal. In the Henderson-Hasselbalch equation, log([base]/[acid]) becomes log(1), which is 0, so pH equals pKa. - 13
A buffer is prepared using 0.40 M lactic acid and 0.04 M lactate ion. The pKa of lactic acid is 3.86. Is the pH higher than, lower than, or equal to 3.86? Explain.
Compare the amount of conjugate base to the amount of weak acid.
The pH is lower than 3.86 because the acid concentration is greater than the conjugate base concentration. In pH = pKa + log([base]/[acid]), the ratio 0.04/0.40 is less than 1, so the logarithm is negative. - 14
Label the main regions on a weak acid strong base titration curve: initial weak acid region, buffer region, half-equivalence point, equivalence point, and excess base region. Describe what happens chemically in the buffer region.
The initial region contains mostly weak acid before much base is added. The buffer region contains both weak acid and conjugate base, so the solution resists pH change. The half-equivalence point is where pH equals pKa. The equivalence point is where all original acid has been neutralized, and the excess base region occurs after extra NaOH has been added. - 15
Compare the equivalence point pH for these two titrations: 0.100 M HCl titrated with 0.100 M NaOH, and 0.100 M CH3COOH titrated with 0.100 M NaOH.
Think about whether the salt formed at equivalence can react with water.
The HCl and NaOH titration has an equivalence point near pH 7 because a strong acid and strong base form a neutral salt solution. The CH3COOH and NaOH titration has an equivalence point above pH 7 because acetate ion, the conjugate base of a weak acid, makes the solution basic.