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Sergeeva-Olga [200]
3 years ago
11

You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How mu

ch solid barium sulfide should you add?
Chemistry
1 answer:
shutvik [7]3 years ago
7 0

Answer:

Explanation:

Molecular weight of barium sulphide = 169

500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide

= .096 moles of barium sulfide

= .096 x 169 gram of barium sulfide

= 16.22 grams of barium sulfide .

We shall have to add 16.22 gram .

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Answer:

pH=4.1

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

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