Question

4. Calculate the pH of a buffer solution made from 0.30 M hydrofluoric acid and 0.70 M sodium fluoride after the addition of 0.04 mol of HCl to 1 L of this solution. Assume no change in volume. K.-7.1x10+ 5. Calculate the pH of a 0.04 M HCl solution. Compare to the pH found in problem 4 Note: this is not a buffer! 6. What properties of a buffer solution provides for a solution with a higher capacity to withstand changes in pH? Provide at least 3.

Answer 1

4.

Moles of HCl added = 0.04 mol

Moles of HF in the buffer solution = 0.30 M x 1 L = 0.30 mol

Moles of NaF in the buffer solution = 0.70 M x 1 L = 0.70 mol

Now, 0.04 mol of HCl react with 0.04 mol of NaF to form 0.04 mol of HF.

NaF + HCl $\rightarrow$ HF + NaCl

Hence, moles of sodium fluoride after the addition of HCl solution = (0.70 - 0.04) mol

                                                                                                         = 0.66 mol

Moles of hydrofluoric acid before the addition of HCl solution = (0.30 + 0.04) mol = 0.34 mol

Molarity of the hydrofluoric acid after the addition of HCl = 0.34 mol/1 L = 0.34 M

Molarity of the sodium fluoride after the addition of HCl= 0.66 mol/1 L = 0.66 M

From Henderson-Hasselbalch equation

pH = pK_a + log([base]/[acid])

      = - logK_a + log([F^-]/[HF])

      = - log(7.1 x 10^-4) + log(0.66/0.34)

     = 3.1 + 0.29

     = 3.4

Hence, the pH of the buffer after the addition of HCl = 3.4

5.

HCl is a strong acid.

Hence, 0.04 M of HCl gives 0.04 M of H₃O⁺ in aqueous solution.

Now,

pH = - log[H₃O⁺]

     = - log(0.04)

     = 1.4

Hence, the concentration of 0.04 M of HCl = 1.4

Now, pH of the buffer after the addition of HCl = 3.4

Hence, pH of the 0.04 M of HCl in buffer solution 2.0 unit higher than the pH of unbuffered 0.04 M of HCl solution .