Question

Your acid has a pka of 7.54 and a concentration of 0.084M.

       You are given 38.6mL of this acid, what is the pH of this initial solution?

       Calculate the pH of the solution after adding 8.7, 17.1, 30.5, and 33.3 mL of 0.100M NaOH.

       At the equivalence point, what volume of base is added and what is the pH?

Answer 1

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An acid with a pKa of 7.54 is a weak acid. To calculate the pH of the initial solution we have to use the equation for Ka:

At equilibrium we can find the concentrations of each of them and the pH by assuming that the weak acid dissociates in an X amount, so there will be X mol of Conjugated base A^- and X mol of H₃O⁺

^{Solving for X we obtain:}

Since the acid is a very weak acid we can simplify the equation by saying that X is a very small portion son C₀ – X is almost equall to C₀. In this case we can find the X with a more simple expression:

Remember that Ka is 10^-pKa or 10^-7.54

The answer for X = 4.9 x 10^-5

The pH is –logX = 4.3

For the next part we are going to need the volume and the amount of NaOH added. The mixture of a weak acid and a strong base will make a buffer for certain range of concentrations. In this case we can use the Henderson-Hasselbalch approximation to calculate the necessary ratio of A- and HA.

We can calculate the concentrations of HA and A^-, we have the pKa and we need the pH. I will make the first quantity as an example because all of them can be determined in the same way.

Number of mol in the original solution: 0.084 x 38.6 mL = 3.2424 mmol

Number of mol of NaOH solution added: 8.7 mL x 0.1M = 0.87 mmol

Since NaOH is a strong base we can confidently say that all of the NaOH react with acid and form conjugated base in the same amount. So [A^-] = 0.87 mmol and [HA] = 3.2424 – 0.87 = 2.3724 mmol

At the equivalence point, what volume of base is added and what is the pH?

At the equivalence point, concentration of the base and the acid are the same. Equivalent point will be achieved when 3.2424 mmol of NaOH had been added. The volume of the base would be:

3.2424 mmol / 0.1M = 32.424 mL

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