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3 years ago

Calculate the pH of a buffer solution that contains 0.25 M benzoic acid (C 6H 5CO 2H) and 0.15M sodium benzoate (C

Chemistry

1 answer:

gtnhenbr [62]3 years ago

4 0

Answer:

$pH=3.97$

Explanation:

Hello,

In this case, for the calculation of the pH of the given buffer we need to use the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa for benzoic acid is 4.19, the concentration of the base is 0.15 M (sodium benzoate) and the concentration of the acid is 0.25 M (benzoic acid), therefore, the pH turns out:

$pH=4.19+log(\frac{0.15M}{0.25M} )\\\\pH=3.97$

Regards.

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Neko [114]

Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

$\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}$

Where:

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Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

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I hope it helps you!

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3 years ago

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