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Basic Buffer Action

40 mL of 0....

Question

$40 m L$ of $0.050 M$ ${N a}_{2} {C O}_{3} + 50 m L$ of $0.040 M$ $H C l$ (write the value to the nearest integer)

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Solution

$m m o l$ of $N a_{2} C O_{3} = 40 \times 0.05 = 2$
$m m o l$ of $H C l = 50 \times 0.04 = 2$
$N a_{2} C O_{3} + H C l ⟶ N a H C O_{3} + N a C l$
$m m o l$ of $N a H C O_{3}$ formed $= 2$
$[N a H C O_{3}] = \frac{2}{40 + 50} = 0.0223$
This is a salt of strong base and weak acid
$p H = 7 + \frac{1}{2} [p K_{a} + log C]$
$⟹ p H = 7 + \frac{1}{2} [6.37 - 1.65]$
$⟹ p H = 9.36$
Nearest integer $= 9$

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