MCQ
$pH$ of $ 0.005\, M\,\,{H_2}S{O_4}$ solution will be
- A$0.005$
- ✓$2$
- C$1$
- D$0.01$
$H _2 SO _{4( aq )} \longrightarrow 2 H ^{+}+ SO _4^{2-}$
Since the equation tells us that each molecule of acid will produce $2$ hydrogen ions, the concentration of the $H ^{+}$ion must be $2 \times 0.0050 \,M$ or $0.010 \,M$
Using the definition of $pH =-\log H ^{+}=-\log (0.010\, M )=2$
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