Can I have some help finding finding the pKa value and mol of HA in this reactio

Question

Can I have some help finding finding the pKa value and mol of HA in this reaction?

Also how do those numbers get plugged into the conjugate base/acid ratio?

How does all of that relate to the mols of OH-?

Strong base is dissolved in 585 mL of 0.400 M weak acid (Ka 4.07 x10) to make a buffer with a pH of 4.14. Assume that the volume remains constant when the base is added. HA(aq)-OH-(aq) H,O(1) A-(aq) Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number HA How many moles of strong base were initially added? Number mol OH

Explanation / Answer

a) PKa = -log Ka = -log (4.07*10-5) = 4.39

Number of mols of HA = Molarity * volume = (0.585 L)(0.400 mol/L) = 0.234 mol

b) Since -

pH = pKa + log [A-]/[HA]

4.14 = 4.39 + log [A-]/[HA]

[A-]/[HA] = 0.562

c) From the above calculation -

   [A-]/0.234 = 0.562

Hence, [A-] = 0.131 mol

If volume chnages are ignorable, the number of mols of base is added initially = 0.131 mol OH-

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