Question

The next three (3) problems deal with the titration of 421 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 2.1 M NaOH.

1. What is the pH of the solution at the 2nd equivalence point?


2. What will the pH of the solution be when 0.1316 L of 2.1 M NaOH are added to the 421 mL of 0.501 M carbonic acid?


3. How many mL of the 2.1 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 6.019?

Answer 1

Question #47454, Chemistry, Inorganic Chemistry

The next three (3) problems deal with the titration of 421 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10⁻⁷, Ka2 = 5.6 x 10⁻¹¹) with 2.1 M NaOH.

1. What is the pH of the solution at the 2nd equivalence point?

2. What will the pH of the solution be when 0.1316 L of 2.1 M NaOH are added to the 421 mL of 0.501 M carbonic acid?

3. How many mL of the 2.1 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 6.019?

ANSWER:

1) pH at the 2nd equal point

$$[ H ^ {+} ] = K w * K a 2 * 2 / C) = \operatorname {s q r t} ((1 0 ^ {\wedge} (- 1 4) ^ {*} 5. 6 ^ {*} 1 0 ^ {\wedge} (- 1 1) ^ {*} 2) / 0. 5 0 1) = 1. 4 9 5 ^ {*} 1 0 ^ {\wedge} (- 1 2)$$ $$\mathrm {p H} = - \lg [ \mathrm {H} + ] = - \lg (1. 4 9 5 * 1 0 ^ {\wedge} (- 1 2)) = \mathbf {1 1 . 8 2}$$

2) If we added 0.1316 L 2.1M NaOH to the 421 mL 0.501 M carbonic acid

$$[ H ^ {+} ] = \sqrt {(K w * K a 1 * \frac {V 1 + V 2}{C * V 1})} =$$ $$= \operatorname {s q r t} ((1 ^ {*} 1 0 ^ {\wedge} (- 1 4) ^ {*} 4. 3 ^ {*} 1 0 ^ {\wedge} (- 7) ^ {*} (0. 1 3 1 6 + 0. 4 2 1)) / (0. 5 0 1 ^ {*} 0. 4 2 1) = 1. 0 6 ^ {*} 1 0 ^ {\wedge} (- 1 0)$$ $$\mathrm {p H} = - \lg (H +) = - \lg (1. 0 6 * 1 0 ^ {\wedge} (- 1 0)) = \mathbf {9}. \mathbf {9 7}$$

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