Editing Answers To More Review Homework For Acid/Base Including Buffers (Non-Math)
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Part I<br> 1. H<sub>2</sub>CO<sub>3</sub>, weak acid<br> 2. Potassium hydroxide, strong base<br> 3. NH<sub>3</sub><br>, weak base 4. S<sub>1</sub>O<sub>3</sub>, non<br> 5. Sulfuric acid, strong acid<br> <br> 6a. pH = 0.00<br> 6b. [OH<sup>1-</sup>] = 9.82E-3 M, pOH = 2.008<br> 6c. pOH = 13.793<br> 7a. 2 K<sub>1</sub>(OH)<sub>1 (aq)</sub> + 1 H<sub>2</sub>SO<sub>4 (aq)</sub> --> 2 H<sub>2</sub>O<sub>1 (l) </sub> + 1 K<sub>2</sub>SO<sub>4 (aq)</sub><br> 7b. [K<sub>1</sub>(OH)<sub>1</sub>] = 2.70M <br> 8. Equivalence point of the solution = Endpoint of the pH indicator<br> 9a. HNO<sub>2</sub><br> 9b. CH<sub>3</sub>NH<sub>3</sub><sup>1+</sup><br> 10a. Weak acid and the salt of its conjugate base<br> <br> 10b. The pH of the solution does not change significantly with the addition of an strong acid or a strong base. It has a specie that will react with the H<sup>1+</sup> so that OH<sup>1-</sup> do not have to be used in this neutralization. Also in the solution is a specie that will do a similar action if an OH<sup>1-</sup> is added to the solution. Remember, pH (and also pOH) scale only keep track of [H<sup>1+</sup>] concentration not acid molecule concentration.<br> <br> <br> <br>
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