How much of 0.3 M ammonium …
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Posted by Prabhnoor Kaur 4 years ago
- 2 answers
Yogita Ingle 4 years ago
Let V mL of NH4OH be mixed with NH4CI to have a buffer of pH 8.65. The total volume after mixing becomes (V+ 30) mL.
m Mole of NH4OH = 0.3×V
Thus, NH4OH = 0.3×V/ (v+30)
m Mole of NH4CL = 0.2×30
Thus, NH4CL = 0.2×30/ (v+30)
= 0.2×30/ (v+30)/ 0.3×V/ (v+30)
= 0.61 = log6 /0.3×V
= v = 4.91mL
Similarly,
14-10 = 4.74+log 0.2×30/ (v+30)/ 0.3×V/ (v+30)
for ph = 10
V = 109.9mL
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Gaurav Seth 4 years ago
How much of 0.3 M ammonium hydroxide should be mixed with 30 ml of 0.2 M solution of ammonium chloride to give buffer solutions 10 pH 8.65 and 10. Given pKb of NH4OH = 4.75.
A n s w e r :
Let V mL of NH4OH be mixed with NH4CI to have a buffer of pH 8.65. The total volume after mixing becomes (V+ 30) mL.
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