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2 years ago

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]2 years ago

6 0

Answer:

205mL of 0.140M KF solution

Explanation:

pH = 2.70 solution.

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

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

Where pH is desire pH = 2.70

pKa is pKa of HF = 3.17

[KF] could be taken as moles of KF

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

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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