The pH of a solution with $[H^+] = 10^{-3}$ M is:
A$10$, since pH = $-\log(10^{-3}) = 10$ in the table
B$11$, mistakenly subtracted from $14$ for OH$^-$ value
C$0.3$, using the natural log instead of base 10 here
D$3$, since pH = $-\log[H^+] = -\log(10^{-3}) = 3$
Answer & Solution
Correct answer: D. $3$, since pH = $-\log[H^+] = -\log(10^{-3}) = 3$
$pH = -\log[H^+] = -\log(10^{-3}) = 3$.
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