Respuesta :

BladeRunner212

The change in pH is 0.04.

Further explanation

Given:

  • 0.50 L of a buffer solution containing 0.15 M in HF and 0.20 M in NaF.
  • The Ka for HF is 3.5 x 10⁻⁴.

Question:

Calculate the change in pH when 2.0 x 10⁻² mol of NaOH is added to this buffer?

The Process:

Step-1

Let us prepare all the moles of substances.

[tex]\boxed{ \ n = MV \ }[/tex]  

Moles of HF =  

[tex]\boxed{ \ 0.15 \ \frac{mol}{L} \times 0.50 \ L = 0.075 \ moles \ }[/tex]

Moles of NaF =  

[tex]\boxed{ \ 0.20 \ \frac{mol}{L} \times 0.50 \ L = 0.10 \ moles \ }[/tex]

Moles of NaOH =

[tex]\boxed{ \ 2.0 \times 10^{-2} = 0.02 \ moles \ }[/tex]

Step-2

Let use the ICE table (in mol).

                        [tex]\boxed{ \ HF + NaOH \rightarrow NaF + H_2O \ }[/tex]

Initial:             0.075     0.02         0.10        -

Change:         0.02       0.02         0.02    0.02

Equlibrium:    0.055       -             0.08    0.02

  • NaOH as a strong base acts as a limiting reagent. The remaining HF as a weak acid and NaF salt forms an acidic buffer system.
  • The NaF salt has valence = 1 according to the number of F⁻ ions as a weak part, i.e.,  [tex]\boxed{ \ NaF \rightleftharpoons Na^+ + F^- \ }[/tex]
  • HF and F⁻ are conjugate acid-base pairs.

Step-3

To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:

[tex]\boxed{ \ pH = pK_a + log\frac{[A^-]}{[HA]} \ }[/tex]

where,  

  • Ka represents the dissociation constant for the weak acid;
  • [A-] represent the concentration of the conjugate base (i.e. salt);  
  • [HA] is the concentration of the weak acid.

[tex]\boxed{ \ pH = pK_a + log\frac{[F^-]}{[HF]} \ }[/tex]

[tex]\boxed{ \ pH = -log(3.5 \times 10^{-4}) + log \Big(\frac{0.08}{0.055}\Big) \ }[/tex]

[tex]\boxed{ \ pH = 4 - log \ 3.5 + 0.1627 \ }[/tex]

[tex]\boxed{ \ pH = 4 - 0.5441 + 0.1627 \ }[/tex]

[tex]\boxed{ \ pH = 3.62 \ }[/tex]

Thus, the buffer pH due to addition of NaOH is equal to 3.62.

Step-4

Let us calculate the buffer pH value before the addition of NaOH.

Moles of HF =  0.075 moles

Moles of NaF = 0.10 mmol

[tex]\boxed{ \ pH = -log(3.5 \times 10^{-4}) + log \Big(\frac{0.10}{0.075}\Big) \ }[/tex]

[tex]\boxed{ \ pH = 4 - log \ 3.5 + 0.1249 \ }[/tex]

[tex]\boxed{ \ pH = 4 - 0.5441 + 0.1249 \ }[/tex]

[tex]\boxed{ \ pH = 3.58 \ }[/tex]

Thus, the initial pH of this buffer is 3.58. This proves the nature of the buffer that keeps the pH value relatively unchanged with the addition of a strong electrolyte.

Finally, we calculate the change in pH.

[tex]\boxed{ \ 3.62 - 3.58 = 0.04 \ }[/tex]

Learn more

  1. The similar problem https://brainly.com/question/9082442
  2. The ratio of HCO₃⁻ to H₂CO₃ in an exhausted marathon runner whose blood pH is 7.2 https://brainly.com/question/3122018
  3. Calculate the pH of an acidic buffer system https://brainly.com/question/9081563

Otras preguntas