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nasty-shy [4]
3 years ago
12

Calculate the [ H + ] and pH of a 0.0040 M butanoic acid solution. The K a of butanoic acid is 1.52 × 10 − 5 . Use the method of

successive approximations in your calculations.
Chemistry
1 answer:
stellarik [79]3 years ago
4 0

Answer:

Ka= [H+][But-]/H[Hbut]

Let k moles of H+ be formed

1.52*10^-5=k^2/0.004-k

K^2=0.004*1.52*10^-5-1.52*10^-5k

K^2- 1.52*10^-5k-6.08*10^-8=0

Solving this quadratic equation.

K=7.6*10^-6

PH=-log(7.6*10^-6)=5.2

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Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

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Now,

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The number of moles of 0.2352 g of potassium hydrogen phosphate

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also,

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Number of moles = 0.2352 / 174.15 = 0.00135 moles

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thus,

for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

or

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3 years ago
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