Calculate the $p H$ of the following mixtures of strong acids, strong bases, and combination of both.
a. $500 m L$ of $0.1 M H C l + 200 m L$ of $0.1 M H_{2} S O_{4} + 300 m L$ of $0.2 M H N O_{3}$ .
b. $100 m L$ of $0.1 M H C l + 100 m L$ of $0.2 M H_{2} S O_{4} + 100 m L$ of $0.1 M H N O_{3}$ and $700 m L$ of $H_{2} O$ .

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Solution

$500 m l$ of $0.1 M$ $H C l$
$200 m l$ of $0.1 M$ $H_{2} S O_{4}$
$300 m l$ of $0.1 M$ $H N O_{3}$
Net Normally $⟹ N = \frac{N_{1} V_{1} + N_{2} V_{2} + N_{3} V_{3}}{V_{1} + V_{2} + V_{3}} = \frac{(0.1) (500) + (0.2) (500) + (0.2) (300)}{1000} = 0.15$
Net Molarity $⟹ M = 0.09$
$p H = - l o g [H^{+}]$
$= - l o g [9 \times 10^{- 2}]$
$= 2 - l o g 3^{2}$
$= 2 - 2 l o g 3$
$= 1.045$

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Calculate the pH of the following mixtures of strong acids, strong bases, and combination of both.a. 500 mL of 0.1 M HCl+200 mL of 0.1 M H2SO4+300 mL of 0.2 M HNO3.b. 100 mL of 0.1 M HCl+100 mL of 0.2 M H2SO4+100 mL of 0.1 M HNO3 and 700 mL of H2O.

500ml of 0.1M HCl200ml of 0.1M H2SO4300ml of 0.1M HNO3Net Normally⟹N=N1V1+N2V2+N3V3V1+V2+V3=(0.1)(500)+(0.2)(500)+(0.2)(300)1000=0.15Net Molarity⟹M=0.09pH=−log[H+] =−log[9×10−2] =2−log32 =2−2log3 =1.045