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Oksanka [162]
3 years ago
14

How many mL (to the nearest mL) of 0.140-M KF solution should be added to 400. mL of 0.212-M HF to prepare a pH

Chemistry
1 answer:
love history [14]3 years ago
6 0

Answer:

205mL of 0.140M KF solution

Explanation:

<em>pH = 2.70 solution.</em>

<em />

It is possible to obtain the pH of the buffer of HF-KF using the H-H equation:

pH = pKa + log [KF] / [HF]

<em>Where pH is desire pH = 2.70</em>

<em>pKa is pKa of HF = 3.17</em>

<em>[KF] could be taken as moles of KF </em>

<em>And [HF] moles of HF: 400.0mL = 0.4L * (0.212mol/L) = 0.0848 moles of HF</em>

<em />

Replacing:

2.70 = 3.17 + log [KF] / [0.0848 moles HF]

-0.47 = log [KF] / [0.0848 moles HF]

0.3388 = [KF] / [0.0848 moles HF]

[KF] = 0.02873 moles of KF must be added.

In mL using concentration of KF (0.140M):

0.02873 moles KF * (1L / 0.140 mol) = 0.205L =

<h3>205mL of 0.140M KF solution</h3>
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