You know what a titration is-we slowly add one solution to another. One solution

Question

You know what a titration is-we slowly add one solution to another. One solution has one of the reactants of chemical reaction; the other solution has the When these is an acid and the other is a base, we have an acid-base titration. We can follow the reaction by measuring the pH of the mixed solution. The result is usually a graph of pH vs. volume added solution. (In lab converted this to mmol of added base, but the principle is the same.) For example, here is from the addition of0.10 M NaOH to 10.0 mL of a containing a weak acid "HA." Your job is to understand what's going on here at all times during this titration. What was the "formal" concentration of HA in the original HA solution? (That is, the molarity you would write on the bottle, not the equilibrium concentration of HA, which would be somewhat less than that.) [This is the easiest question-all you need is equivalence point.] What is the Ka of this acid? [This is the second-earliest question to answer, I'll give you three choices 1.3 times 10^-3, 1.8 times 10^-5, or 1.8 times 10^-9.) What are the amounts or concentrations of the principal species in solution initially? After 2 mL are added? 5 mL? 10 mL? 15 mL? [This is just accounting.] Using the pKa and formal concentration, calculate the predicted pH anywhere along this curve. This is more involved. It is a different calculation depending upon what point we are talking about in the titration. The graph above is your answer key.

Explanation / Answer

(1) Concentration of weak acid sholud be 0.10 M only because it requires 10.0 mL of 0.10 M NaOH to reach equivalence point.

(2) From the graph, the equivalence point is occured at pH = 8.74 (around) . And at equivalence point pH = pKa.

so, Ka = 10-8.74 = 1.8 * 10-9

(3) At 0 mL, HA, H3O+, A-

At 2mL , HA, A-, Na+

At 5mL HA, A-, Na+

at 10mL A-, Na+, OH-

At 15 mL: A-, Na+, OH-, HA

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