Chemistry, 30.09.2019 16:00, realmofgrads23

Calculate the change in ph when 2.0×10−2mol of naoh is added to 0.50 l of a buffer solution that is 0.15 m in hf and 0.20 m in naf.

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answered: YSGabino6019

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.

$\boxed{ \ n = MV \ }$

Moles of HF =

$\boxed{ \ 0.15 \ \frac{mol}{L} \times 0.50 \ L = 0.075 \ moles \ }$

Moles of NaF =

$\boxed{ \ 0.20 \ \frac{mol}{L} \times 0.50 \ L = 0.10 \ moles \ }$

Moles of NaOH =

$\boxed{ \ 2.0 \times 10^{-2} = 0.02 \ moles \ }$

Step-2

Let use the ICE table (in mol).

$\boxed{ \ HF + NaOH \rightarrow NaF + H_2O \ }$

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., $\boxed{ \ NaF \rightleftharpoons Na^+ + F^- \ }$ 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:

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

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.

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

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

$\boxed{ \ pH = 4 - log \ 3.5 + 0.1627 \ }$

$\boxed{ \ pH = 4 - 0.5441 + 0.1627 \ }$

$\boxed{ \ pH = 3.62 \ }$

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

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

$\boxed{ \ pH = 4 - log \ 3.5 + 0.1249 \ }$

$\boxed{ \ pH = 4 - 0.5441 + 0.1249 \ }$

$\boxed{ \ pH = 3.58 \ }$

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.

$\boxed{ \ 3.62 - 3.58 = 0.04 \ }$

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