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Annette [7]
3 years ago
11

A buffer solution contains 0.368 M hydrocyanic acid and 0.360 M potassium cyanide . If 0.0513 moles of sodium hydroxide are adde

d to 225 mL of this buffer, what is the pH of the resulting solution
Chemistry
1 answer:
MrRa [10]3 years ago
7 0

Solution :

Millimoles of hydrocyanic acid = 225 x 0.368

                                                   = 82.8

Millimoles of potassium cyanide = 225 x 0.360

                                                   = 81

Millimoles of sodium hydroxide = 51.3

Therefore,

pOH = pKb + log [salt - C / bas + C]

        = 4.74 + log[82.8 - 51.3 / 81 + 51.3]

         = 4.102

Therefore, pH = 9.05

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0.116 grams

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Answer:

Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

Explanation:

Mass percentage -

Mass percentage of A is given as , the mass of the substance A by mass of the total solution multiplied by 100.

i.e.

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Given,  

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1. Ammonia, NH₃

Hence , the mass of solution is calculated as the summation of the mass of the atom * number of atom ,  

Hence ,  

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The mass percent of nitrogen can be calculated from using the above formula  

mass % N = mass of N / mass of solution * 100

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mass % N = mass of N / mass of solution * 100

mass % N = 14 g/mol / 80 g/mol * 100

mass % N = 17.50 % .

3.   ammonium hydrogen phosphate, (NH₄)₂HPO₄

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mass of solution = 2 * 14 g/mol  + 9 * 1 g/mol  + 4 * 16 g/mol + 1 * 31 g/mol

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Ammonia, NH₃ > ammonium nitrate, NH₄NO₃ > ammonium hydrogen phosphate, (NH₄)₂HPO₄

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