Ionic EquilibriumHard
Question
What is the pH of a neutral solution at 37°C, where Kw equals 2.5 × 10–14? (log 2 = 0.3)
Options
A.7.0
B.13.6
C.6.8
D.6.6
Solution
$\left\lbrack H^{+} \right\rbrack = \sqrt{K_{w}} \Rightarrow P^{H} = - {\log\left( 2.5 \times 10^{- 14} \right)}^{1/2} = 6.8$
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