Calculate the pH for -
$50 m L o f 0.1 M H C l . 25 m L o f 0.1 M H_{2} S O_{4} . 25 m L o f 0.2 M H N O_{2} + 100 m L o f H_{2} O$

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Solution

$(N_{1} V_{1})_{H C l} + (N_{2} V_{2})_{H_{2} S O} + (N_{3} V_{3})_{H N O_{3}} = N (V_{1} + V_{2} + V_{3} + 180)$
$0.1 \times 50 + 2 (0.1) \times 25 + (1) (25) (0.2) = N (50 + 25 + 25 + 100)$
$N = [H^{+}) = \frac{15}{200} = 0.075$
$p H = - l o g (0.075)$
= 1.12

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