Calculate the pH of orange juice in which hydronium ion conc. $(H_{3} O^{+})$ is 2 $\times$ $10^{- 8}$ $m o l / l$ .

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Solution

$C o n c e n t r a t i o n o f H y d r o n i u m i o n (H_{3} O^{+}) = 2 \times 10^{- 8} m o l e s / i$

$[H_{3} O^{+}] = 2 \times 10^{- 8}$

$p H = - log [H_{3} O^{+}]$

$p H = - log [2 \times 10^{- 8}]$

$= 7.69$

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