What would be the pH at the half-equivalence point of the titration of an acid w

Question

What would be the pH at the half-equivalence point of the titration of an acid with a K_a of 6.3 times 10^-6? A solution is prepared by adding equal numbers of moles of sodium acetate (NaC_2H_3O_2) and HC_2H_3O_2. What is the pH of the solution? A student standardizing her NaOH solution for this experiment dissolved 0.153 g of KHP in 50.0 mL of water and then titrated the acid. How would the results of the standardization have been affected if the KHP had been dissolved in 100.0 mL of water, instead of 50.0 mL? How many milliliters of 0.100M NaOH solution would be required to titrate 0.250 g of butyric acid (MM: 88.1 g/mol) to the equivalence point? What would the mixture pH be at the half-equivalence point? Explain the effect each of the following procedural errors would have on the calculated morality of the NaOH solution. as determined using the Procedure of this experiment. The buret tip was not filled with NaOH solution before the initial buret reading was taken. Some of the titration mixture splashed out of the flask during the titration. The buret was rinsed with water, not NaOH solution, before it was filled with NaOH solution for titration of the KHP. How would the calculated equivalent mass of the unknown add in this experiment be affected If: The buret was contaminated with acid solution? Not all of the unknown acid was transferred from the weighing paper to the titration beaker? The volume of distilled water in which the unknown acid was dissolved was not accurately known?

Explanation / Answer

pH = pKa + log [base]/[acid]
where "base" and "acid" are the weak acid and its conjugate base.

At half equivalence point, the concentrations of the weak acid and its conjugate base are equal. So that ratio is 1.

The log of 1 is zero, so, the pH = pKa.

Hence pH= pKa = -log(6.3*10-8 +10^-7) = 6.7878

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