What is the $p H$ $pH$ of a $0.055 \cdot m o l \cdot L^{- 1}$ $0.055molL^-1$ solution of sulfuric acid in water?

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Answer 1 · 2016-05-29T19:25:51

Assuming quantitative dissociation, $p H = 0.96$ $pH=0.96$

We assume (reasonably!) that dissociation is quantitative:

$H_{2} S O_{4} (a q) + 2 H_{2} O \to S O_{4}^{2 -} + 2 H_{3} O^{+}$ $H_2SO_4(aq) + 2H_2O rarr SO_4^(2-) + 2H_3O^+$

Thus $[H_{3} O^{+}]$ $[H_3O^+]$ $=$ $=$ $0.110 \cdot m o l \cdot L^{- 1}$ $0.110*mol*L^-1$ .

Now $p H$ $pH$ $=$ $=$ $- {log}_{10} [H_{3} O^{+}]$ $-log_10[H_3O^+]$ $=$ $=$ $- {log}_{10} (0.110)$ $-log_10(0.110)$ $=$ $=$ $0.96$ $0.96$

What is the pHpHpH of a 0.055*mol*L^-10.055⋅mol⋅L−10.055*mol*L^-1 solution of sulfuric acid in water?