Calculate the pH of the solution when the following amounts of NaOH have been added

Question

A 15.00 mL sample of 0.500 molar aceic acid is being titrated with 0.750 molar sodium hydroxide. The Ka of acetic acid is 1.76 x 10^ -5.

 

(a) 0.00 mL

 

 

(b) 2.50 mL

 

(c) 5.00 mL

 

I need some serious help with this. I don't even know where to start.

Answer 1

Hello Hope,

 

In this problem you are titrating a weak acid (acetic acid, CH₃COOH) with a strong base (sodium hydroxide, NaOH).  The acid ionization constant (Ka) for acetic acid  is given as 1.76x10⁵.  The -log[Ka] = pKa = 4.75.   Before the titration begins, you need to calculate the pH of the acid as follows.

 

HAc        ↔        H⁺ + Ac^-

0.500 M             0       0  ⇐ Initial

-x                     +x    +x ⇐ change

0.500 M-x            x      x ⇐ equilibrium

 

Ka = [H⁺][Ac^-]/[HAc]

 

1.76x10⁵ = [x][x]/[0.500-x] ⇒ Ka < 10^-3 so we may ignore the x in the denominator to avoid a quadratic.

1.76x10⁵ = [x2]/[0.500]

x = 2.9665x10^-3 = [H^+]

pH = -log[H+] = -log[2.9665x10^-3] = 2.73

 

 

Now we titrate the acetic acid with 2.50 mL of 0.750 M NaOH.

 

HAc         +         NaOH →         NaAc + H₂0

15.00 mL            2.50 mL

0.500 M              0.750 M

Calculate moles:       0.00750 mol      0.001875 mol        0

Subtract moles of limiting reactant:  -0.001875 mol   -0.001875 mol   +0.001875 mol ⇐ 0.001875 mol of the salt form.

At equilibrium:         0.0056225 mol         0 mol           0.001875 mol

 

 

At this point in the titration, we have a buffer solution consisting of the weak acid and its conjugate base (HAc and Ac^-).  To calculate the pH of a buffer, use the Henderson-Hasselbalch equation.

 

pH = pKa + log[conjugate base/acid] = 4.75 +log[0.001875 mol/0.005625 mol] = 4.28

 

We usually take the log₁₀ of molar concentrations, but dividing the mole values above by the same volume (in liters) will yield exactly the same answer.

 

 

Now we titrate the acetic acid with 5.00 mL of 0.750 M NaOH.

 

Doing exactly what I did for 2.50 mL of NaOH above yields 0.00375 moles of Ac, 0 moles of NaOH, and 0.00375 moles of NaAc.  When the moles of the weak acid equal the moles of its conjugate base, we are at the half-titration point, and pH = pKa.

 

pH = 4.75 + log[0.00375/0.00375] = 4.75 ⇐ The log₁₀ of one equals zero.

 

All the best,

Randall