Question

0.3931 0.34:1 12:1 D Question 5 5 pts A 25.0 mL sample of 0.150 M hydrofluoric acid is titrated with a 0.150 M NaOH solution. What is the concentration of [F 1. Kb.[OH formed from the weak base ionization, and pH at the equivalence point? The K, of hydrofluoric acid is 3.5 x 10 In the space below enter your numerical answers in order of 1 to 4 include units. 1- Molarity of the Fluoride ion formed 2. The numerical value for kb. 3. Concentration of hydroxide ion 4. pH of the solution formed. Equation Sheet BIVAA-I EI 311 XX, EE - ? VBG 1: 12pt - Paragraph

Answer 1

5.

Weak acid HF ionizes to give

$HF\rightarrow H^{+}+F^{-}$

As per this balanced equation, concentration of H+ is equal to concentraion of F-.

Ionization constant of weak acid HF $K_{a}=\frac{[H^{+}][F^{-}]}{[HF]}$

Given that Ka = 3.5 x 10-4

3.5 x 10-4-1 -4 [H+1F-1 HF

As H+ and F- concentrations are equal

  $3.5\times 10^{-4}=\frac{[F^{-}]^{2}}{[HF]}$

$3.5\times 10^{-4}=\frac{[F^{-}]^{2}}{0.150}$

  ${[F^{-}]^{2}}=0.150\times 3.5\times 10^{-4}=0.525\times 10^{-4}$

${[F^{-}]}=\sqrt{0.525\times 10^{-4}}=0.725\times 10^{-2}M$

Ka value of HF is given as 3.5 x 10-4. But we know that

$K_{a}K_{b}=10^{-14}$

  $K_{b}=\frac{10^{-14}}{3.5\times 10^{-4}}=2.86\times 10^{-11}$

In the solution of HF, [H+] = [F-] = 0.725 x 10-2 M

We know that, in an aqueous solution $[H^{+}][OH^{-}]=10^{-14}$

$[OH^{-}]=\frac{10^{-14}}{[H^{+}]}=\frac{10^{-14}}{0.725\times 10^{-2}}=1.379\times 10^{-12}M$

At equivalence point weak acid HF and strong base NaOH are completely neutralized to give salt NaF.

pH of solution of a salt of weak acid and weak base can be determined using the following formula

  $pH=7+\frac{1}{2}(pK_{a}+logC)$ C is the concentration of salt solution

At equivalence point, 25 mL 0.150 M HF reacts with 25 mL 0.150 M NaOH and gives the salt solution of NaF.

As equal volumes of acid and base solutions are mixed, the volume of the salt solution becomes 50 mL.

Volume of the solution has become double. So, concentration salt solution will be half to expected.

Concentraton of aqueous solution of NaF = 0.150 / 2 = 0.075 M

pKa can be determined using the formula, pKa = - log Ka

pKa = - log (3.5 x 10-4) = 4 - log 3.5 = 4 - 0.544 = 3.456

Now pH of the salt solution

  $pH=7+\frac{1}{2}(3.456+log0.075)=7+\frac{1}{2}(3.456-1.1249)=7+1.1656=8.1656$