Calculate the pH after 0.015 mole of HCl is added to 1.00 L of each of the four solutions. (Assume that all solutions are at 25°C.)

Question

(a)    0.141 M pentanoic acid (HC₅H₉O₂, K_a = 1.5 ✕ 10⁻⁵)

(b)    0.141 M sodium pentanoate (NaC₅H₉O₂)

(c)    pure H₂O

(d)    0.141 M HC₅H₉O₂ and 0.141 M NaC₅H₉O₂

Please help. And please show the steps. I am quite confused on how to find the Pka and with using the Henderson Hasselbalch equations. This is what I have gotten so far. 

(a)

Ka=1.5x10^-5 = [H^+][A-]/0.141=x^2/0.141

x^2=2.115e-6

x=0.001454304 M = [H^+]

(c)

pH=-log(0.015)=1.82

Answer 1

Before doing the actual problem, let us look at pKa so that you can use the HH equation efficiently. Of course, since the HH equation is essentially derived from the Ka expression, you don't have to use HH, but it is much easier. So ....

pKa = -log Ka (just as pH is the negative log of H+). So, if the Ka for pntanoic acid is 1.5x10^-5, the pKa will be the negative log of that or 4.82.

Now, the HH equation is pH = pKa + log [salt]/[acid] or pOH = pKb + log [salt]/[base]. Let us look at question (a). adding 0.015 mol HCl to 1.00 L of 0.141 M pentanoic acid, HC₅H₉O₂. Here you are simply adding a strong acid to a weak acid. THERE IS NO BUFFER. Essentially, the pH of this solution will be dictated by the HCl with very little effect coming from the weak acid. Thus pH = -log 0.015 = 1.8

(b). adding 0.015 mol HCl to 1.00 L of 0.141 M sodium pentanoate. Addition of HCl will result in formation of NaCl and the weak acid, pentanoic acid. The reaction looks like this: HCl + NaC₅H₉O₂ = NaCl + HC₅H₉O₂ and depending on moles of HCl and sodium pentanoate, we may form a buffer. Let's look.

HCl + NaC₅H₉O₂ = NaCl + HC₅H₉O₂

0.015.......0.141..........0...............0...........Initial

-0.015.....-0.015........+0.015....+0.015.....Change

0...........0.126..........0.015........0.015......Equilibrium

So, we have formed a buffer with 0.015 M pentanoic acid + 0.126 M sodium pentanoate (weak acid + salt of the weak acid = buffer). Using the HH equation we can substitute our values from the ICE table above.

pH = pKa + log [salt]/[acid]

pH = 4.82 + log [0.126]/[0.015]

pH = 4.82 + log 8.4 = 4.82 + 0.92

pH 5.74

(c). Yes, that is correct.

(d). adding 0.015 moles HCl to 1.00 L of equimolar acid and salt. HCl will react with the pentanoate anion as follows:

H⁺ + C₅H₉O₂^- ===> HC₅H₉O₂ thus forming more of the weak acid. Lookin at the ICE table....

0.015....0.141.................0.141..........Initial

-0.015...-0.015............+0.015...........Change

0..........0.126...............0.156...........Equililbrium

Using HH equation and plugging in the values from the ICE table, we have...

pH = 4.82 + log [0.126]/[0.156]

pH = 4.82 + log 0.808 = 4.82 + (-0.09)

pH = 4.82 - 0.09 = 4.73