The ph of solution obtained by mixing 500ml of 0.15m h2so4 with 0.1m naoh is
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Answer:M1OHH2O⇌M1++OH−
0.4−x x x
Kb1=0.4−xx2
since x is small
Kb1=0.4x2
x2=0.4Kb1
x2=4×10−8
x=2×10−4
[OH−]=2×10−4
N2OHH2O⇌N2++OH−
30.8−y y y
Kb2=
Explanation:
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