Q: A 25.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. What is the pH after 10.0 mL of NaOH has been added? (Assume volumes are additive.)

$$\text{First, calculate moles of HCl:} \\ 0.0250~\text{L} \times 0.100~\text{mol/L} = 0.00250~\text{mol} \\ \text{Moles of NaOH added:} \\ 0.0100~\text{L} \times 0.100~\text{mol/L} = 0.00100~\text{mol} \\ \text{Moles of HCl remaining:} \\ 0.00250 - 0.00100 = 0.00150~\text{mol} \\ \text{Total volume:} \\ 25.0~\text{mL} + 10.0~\text{mL} = 35.0~\text{mL} = 0.0350~\text{L} \\ [\text{H}^+] = \frac{0.00150}{0.0350} = 0.0429~\text{M} \\ \text{pH} = -\log(0.0429) = 1.37$$

Q: During the titration of 50.0 mL of 0.100 M acetic acid ($$\mathrm{p}K_a = 4.76) $$with 0.100 M NaOH, what is the pH when 25.0 mL of NaOH has been added?

$$\text{At half-equivalence, } \text{pH} = \text{p}K_a = 4.76$$

Q: Which equation would you use to calculate the pH of a buffer solution formed during the titration of a weak acid with a strong base, before the equivalence point?

$$\text{Henderson-Hasselbalch equation:} \\ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$$

Q: A 40.0 mL sample of 0.200 M HNO$$_3 is $$titrated with 0.100 M KOH. What volume of KOH is required to reach the equivalence point?

$$\text{Moles of HNO}_3 = 0.0400~\text{L} \times 0.200~\text{mol/L} = 0.00800~\text{mol} \\ \text{Moles of KOH needed} = 0.00800~\text{mol} \\ \text{Volume of KOH} = \frac{0.00800~\text{mol}}{0.100~\text{mol/L}} = 0.0800~\text{L} = 80.0~\text{mL}$$

Q: A 25.0 mL sample of 0.100 M ammonia ($$\mathrm{NH}_3$$, $$K_b = 1.8 \times 10$$^{-5}$$) is titrated with 0.100 M HCl. What is the pH at the equivalence point?

$$\text{At equivalence, solution contains NH}_4^+ \text{ only.} \\ \text{NH}_4^+ + \text{H}_2\text{O} \rightarrow \text{NH}_3 + \text{H}_3\text{O}^+ \\ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10$$^{-14}$$}{1.8 \times 10$$^{-5}$$} = 5.56 \times 10$$^{-10}$$ \\ [\text{NH}_4^+] = 0.0500~\text{mol/L} \\ \text{Set up ICE table and solve:} \\ x^2$/(0.0500) = 5.56 \times 10$$^{-10}$$ \\ x = [\text{H}_3\text{O}^+] = 5.27 \times 10$$^{-6}$$ \\ \text{pH} = 5.28$$

Question 1

A 25.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. What is the pH after 10.0 mL of NaOH has been added? (Assume volumes are additive.)

Accepted Answer

$$	ext{First, calculate moles of HCl:} \ 0.0250~	ext{L} 	imes 0.100~	ext{mol/L} = 0.00250~	ext{mol} \ 	ext{Moles of NaOH added:} \ 0.0100~	ext{L} 	imes 0.100~	ext{mol/L} = 0.00100~	ext{mol} \ 	ext{Moles of HCl remaining:} \ 0.00250 - 0.00100 = 0.00150~	ext{mol} \ 	ext{Total volume:} \ 25.0~	ext{mL} + 10.0~	ext{mL} = 35.0~	ext{mL} = 0.0350~	ext{L} \ [	ext{H}^+] = \frac{0.00150}{0.0350} = 0.0429~	ext{M} \ 	ext{pH} = -\log(0.0429) = 1.37$$

Question 2

During the titration of 50.0 mL of 0.100 M acetic acid ($$\mathrm{p}K_a = 4.76) $$with 0.100 M NaOH, what is the pH when 25.0 mL of NaOH has been added?

Accepted Answer

$$	ext{At half-equivalence, } 	ext{pH} = 	ext{p}K_a = 4.76$$

Question 3

Which of the following best describes the pH at the equivalence point when titrating a weak acid with a strong base?

Accepted Answer

The pH is greater than 7 because the conjugate base hydrolyzes to produce $$\mathrm{OH}^-.$$

Question 4

A 0.050 M solution of sulfuric acid ($$\mathrm{H}_2\mathrm{SO}_4) is $$titrated with 0.100 M NaOH. Which of the following statements is true about the titration curve?

Accepted Answer

There are two equivalence points, corresponding to the two acidic protons of $$\mathrm{H}_2\mathrm{SO}_4.$$

Question 5

Which equation would you use to calculate the pH of a buffer solution formed during the titration of a weak acid with a strong base, before the equivalence point?

Accepted Answer

$$	ext{Henderson-Hasselbalch equation:} \ 	ext{pH} = 	ext{p}K_a + \log \left( \frac{[	ext{A}^-]}{[	ext{HA}]} ight)$$

Question 6

A 40.0 mL sample of 0.200 M HNO$$_3 is $$titrated with 0.100 M KOH. What volume of KOH is required to reach the equivalence point?

Accepted Answer

$$	ext{Moles of HNO}_3 = 0.0400~	ext{L} 	imes 0.200~	ext{mol/L} = 0.00800~	ext{mol} \ 	ext{Moles of KOH needed} = 0.00800~	ext{mol} \ 	ext{Volume of KOH} = \frac{0.00800~	ext{mol}}{0.100~	ext{mol/L}} = 0.0800~	ext{L} = 80.0~	ext{mL}$$

Question 7

At what point during the titration of a weak acid with a strong base does $$	ext{pH} = 	ext{p}K_a$$?

Accepted Answer

When half of the acid has been neutralized (half-equivalence point).

Question 8

A 25.0 mL sample of 0.100 M ammonia ($$\mathrm{NH}_3$$, $$K_b = 1.8 	imes 10$$^{-5}$$) is titrated with 0.100 M HCl. What is the pH at the equivalence point?

Accepted Answer

$$	ext{At equivalence, solution contains NH}_4^+ 	ext{ only.} \ 	ext{NH}_4^+ + 	ext{H}_2	ext{O} ightarrow 	ext{NH}_3 + 	ext{H}_3	ext{O}^+ \ K_a = \frac{K_w}{K_b} = \frac{1.0 	imes 10$$^{-14}$$}{1.8 	imes 10$$^{-5}$$} = 5.56 	imes 10$$^{-10}$$ \ [	ext{NH}_4^+] = 0.0500~	ext{mol/L} \ 	ext{Set up ICE table and solve:} \ x^2$/(0.0500) = 5.56 	imes 10$$^{-10}$$ \ x = [	ext{H}_3	ext{O}^+] = 5.27 	imes 10$$^{-6}$$ \ 	ext{pH} = 5.28$$

Question 9

Which of the following best describes the buffer region in a titration curve of a weak acid with a strong base?

Accepted Answer

The region before the equivalence point where both weak acid and its conjugate base are present in significant amounts.

Question 10

A 0.100 M solution of phosphoric acid ($$\mathrm{H}_3\mathrm{PO}_4) is $$titrated with 0.100 M NaOH. How many equivalence points will be observed, and why?

Accepted Answer

Three equivalence points, because $$\mathrm{H}_3\mathrm{PO}_4 is a $$triprotic acid.

Question 11

Which of the following is true about the titration of a strong acid with a strong base?

Accepted Answer

The pH at the equivalence point is 7.00 at 25$$^\circC.$$

Question 12

A 50.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. What is the pH after 60.0 mL of NaOH has been added?

Accepted Answer

$$	ext{Moles HCl} = 0.0500~	ext{L} 	imes 0.100~	ext{mol/L} = 0.00500~	ext{mol} \ 	ext{Moles NaOH} = 0.0600~	ext{L} 	imes 0.100~	ext{mol/L} = 0.00600~	ext{mol} \ 	ext{Excess NaOH:} 0.00600 - 0.00500 = 0.00100~	ext{mol} \ 	ext{Total volume:} 50.0~	ext{mL} + 60.0~	ext{mL} = 110.0~	ext{mL} = 0.110~	ext{L} \ [	ext{OH}^-] = \frac{0.00100}{0.110} = 0.00909~	ext{M} \ 	ext{pOH} = -\log(0.00909) = 2.04 \ 	ext{pH} = 14.00 - 2.04 = 11.96$$

Question 13

Which of the following best explains why the pH changes rapidly near the equivalence point in a titration curve?

Accepted Answer

The buffering capacity is minimal, so small additions of titrant cause large changes in pH.

Question 14

A 25.0 mL sample of 0.100 M oxalic acid ($$\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4) is $$titrated with 0.100 M NaOH. How many moles of NaOH are required to reach the second equivalence point?

Accepted Answer

$$	ext{Oxalic acid is diprotic:} \ 	ext{Moles of acid:} 0.0250~	ext{L} 	imes 0.100~	ext{mol/L} = 0.00250~	ext{mol} \ 	ext{Each mole requires 2 moles NaOH:} \ 0.00250~	ext{mol} 	imes 2 = 0.00500~	ext{mol}$$

Question 15

Which of the following is a correct interpretation of the point where $$	ext{pH} = 	ext{p}K_a$$ during a titration?

Accepted Answer

The concentrations of weak acid and its conjugate base are equal.

Question 16

A 100.0 mL sample of 0.050 M citric acid ($$\mathrm{H}_3\mathrm{C}_6\mathrm{H}_5\mathrm{O}_7) is $$titrated with 0.100 M NaOH. How many mL of NaOH are required to reach the first equivalence point?

Accepted Answer

$$	ext{Moles citric acid:} 0.100~	ext{L} 	imes 0.050~	ext{mol/L} = 0.00500~	ext{mol} \ 	ext{First equivalence: 1 mol NaOH per mol acid} \ 	ext{Volume NaOH:} \frac{0.00500~	ext{mol}}{0.100~	ext{mol/L}} = 0.0500~	ext{L} = 50.0~	ext{mL}$$

Question 17

Which of the following best describes the pH at the equivalence point when titrating a polyprotic acid with a strong base?

Accepted Answer

The pH at each equivalence point depends on the strength of the conjugate base formed at that stage.

Acid-Base Titrations: Calculations and Interpretation

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