the degree of ionisation of 0.5N NH3 at 25 degree centigrade in a solution of ph is 12
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3
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answer to your question is 2%
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1
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Solution :
NH4OHCmol L−1C−Cα⇔NH+40Cα+OH−0Cα
pH = 12 means [H+]=10−12or[OH−]=10−2
∴[OH+]=Cα=10−2orα=10−2C=10−20.5=2×10−2or2%.
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