A buffer solution is prepared by mixing 56.2 mL of 0.0354 M ammonium chloride wi
ID: 918312 • Letter: A
Question
A buffer solution is prepared by mixing 56.2 mL of 0.0354 M ammonium chloride with 76.3 mL of 0.749 M ammonia.
A table of pKa values can be found here.
1. Calculate the pH (to two decimal places) of this solution.
Assume the 5% approximation is valid and that the volumes are additive.
2. Calculate the pH (to two decimal places) of the buffer solution after the addition of 6.57 g of ammonia (NH3) to the buffer solution above.
Assume 5% approximation is valid and that the volume of solution does not change.
A buffer solution is prepared by mixing 92.0 mL of 0.466 M carbonic acid with 20.5 mL of 0.564 M sodium hydrogen carbonate.
A table of pKa values can be found here.
1. Calculate the pH (to two decimal places) of this solution.
Assume the 5% approximation is valid and that the volumes are additive.
2. Calculate the pH (to two decimal places) of the buffer solution after the addition of 194 mL of a 0.00337 M solution of hydroiodic acid to the existing buffer solution.
Assume the 5% approximation is valid and that the volumes are additive.
Calculate the buffer ratio (base/acid) required for a buffer of pOH = 8.32 that is prepared by mixing propanoic acid and sodium propanoate. A table of pKa values can be found here. Report your answer to 2 significant figures in scientific notation.
Explanation / Answer
there are 3 long question each with two parts so, i am solving the 1 question completely: please post remaining question sepapartely so that it can be solved properly.
A buffer solution is prepared by mixing 56.2 mL of 0.0354 M ammonium chloride with 76.3 mL of 0.749 M ammonia.
A table of pKa values can be found here.
1. Calculate the pH (to two decimal places) of this solution.
Assume the 5% approximation is valid and that the volumes are additive.
2. Calculate the pH (to two decimal places) of the buffer solution after the addition of 6.57 g of ammonia (NH3) to the buffer solution above.
SOlution :
1) pKb = 4.74 for NH3 , Its a basic buffer so
pOH = pKb + log(salt/base)
= 4.74 + log(56.2 mL x 0.0354 M / 76.3 mL x0.749 M )
=4.74 -1.458 =3.28 1
pH = 14 - pOH = 14 -3.281 = 10.718 answer
2) on addition of NH3 , The millimoles of Ammonia increases
so Total Millimoles of NH3 = 76.3 mL x0.749 M + (6.57/17)1000= 443.61
pOH = pKb + log(salt/base)
= 4.74 + log(56.2 mL x 0.0354 M / 443.61 )
=4.74 -2.34 =2.39
pH = 14 - pOH = 14 -2.39 = 11.608 answer
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