You are titrating an unknown amount of butanoic acid (CH_3CH_2CH_2CO_2H, Ka = 1.

Question

You are titrating an unknown amount of butanoic acid (CH_3CH_2CH_2CO_2H, Ka = 1.52 times 106-5), with 0.1 M NaOH. 25 mL of the unknown required 37.5 mL of titrant (the 0.1 M NaOH). What is the concentration of the unknown? Calculate the pH of the solution at the endpoint. Now I'm going to break this into steps for you: First determine the concentration of the butanoic acid. (This is similar to your lab final.) Now we will use that to determine the concentration of the conjugate base. Here is the reaction that happened in the flask: CH_3CH_2CH_2CO_2H + NaOH CH_3CH_2CH_2CO_2Na + H_2O To find the pH at the END POINT we need to know the concentration of the underlined species (conjugate base). It is not the same as the concentration of the unknown, because we added water (37.5 mL of titrant) to the flask. So take the moles of the unknown divide by the new total volume (25 + 37.5 mL) to get the concentration at the end point. Now we have a weak base problem. Convert Ka to Kb and use the concentration you got at the end point and do a normal ICE table (remember the times is [OH^-]) and convert to pH.

Explanation / Answer

CH3CH2CH2CO2H + NaOH <=====> CH3CH2CH2CO2Na + H2O

At the equivalence point,

1 mole CH3CH2CH2CO2H = 1 mole NaOH

Moles of NaOH = (37.5 mL)*(1 L/1000 mL)*(0.1 mole/L) = 0.00375 mole.

Therefore, moles of CH3CH2CH2CO2H = 0.00375 mole.

Concentration of CH3CH2CH2CO2H = (0.00375 mole)/[(25 mL)*(1 L/1000 mL)] = 0.15 mol/L = 0.15 M (ans).

At the equivalence point, we have 0.15 M, we have 0.00375 mole CH3CH2CO2H neutralized and we have 0.00375 mole CH3CH2CH2CO2Na.

Volume of the solution = (25 + 37.5) mL = 62.5 mL = (62.5 mL)*(1 L/1000 mL) = 0.0625 L.

Concentration of CH3CH2CH2CO2Na = (0.00375 mole)/(0.0625 L) = 0.06 mol/L = 0.06 M (ans).

We have the Ka of CH3CH2CH2CO2H given; we can find the Kb as

Kb = Kw/Ka = (1.0*10-14)/(1.52*10-5) = 6.5789*10-10

Consider the dissociation of CH3CH2CH2CO2Na as below:

CH3CH2CH2CO2Na + H2O <=====> CH3CH2CH2CO2H + OH-

Since OH- is formed, work with Kb.

Kb = [CH3CH2CH2CO2H][OH-]/[CH3CH2CH2CO2Na]

===> 6.5789*10-10 = (x).(x)/(0.06 – x)

Use the small x approximation; since Kb is small, we can assume x << 0.06, we must have

6.5789*10-10 = x2/0.06

===> 3.94734*10-11 = x2

===> x = 6.2828*10-6

Therefore, [OH-] = 6.2828*10-6 M and pOH = -log [OH-] = -log (6.2828*10-10) = 9.2018

Thus, pH = 14 – pOH = 14 – 9.2018 = 4.7982 4.80 (ans).

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