Ionic EquilibriumHard
Question
At 90o C, the hydronium ion concentration in pure water is 10−6 M. If 100 ml of 0.5 M-NaOH solution is mixed with 250 ml of 0.2 M-HNO3 solution at 90o C, then pH of the resulting solution will be
Options
A.7.0
B.6.0
C.8.0
D.0.85
Solution
$n_{OH^{-}}\text{ taken = }\frac{100 \times 0.5}{1000} = 0.05$
$n_{H^{+}}\text{ taken = }\frac{250 \times 0.2}{1000} = 0.05$
Hence, resulting solution is neutral PH = –log(10–6) = 6.0
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