$H_{3} P O_{4}$ is a tri basic acid and one of its salt is $N a H_{2} P O_{4}$ . What volume in ml of $1$ M NaOH solution should be added to $12$ g of $N a H_{2} P O_{4}$ to convert it into $N a_{3} P O_{4}$ ?

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Solution

1) $N a H_{2} P O_{4} + 2 N a O H \to N a_{3} P O_{4} + H_{2} O$
2) Thus according to this reaction $1 m$ of $N a H_{2} P O_{4}$ react with $2$ mol of $N a O H$ .
Molar mass of $N a H_{2} P O_{4} = 120 g / m o l$
Number of moles of $N a H_{2} P O_{4}$ in $12 g = 0.1 m o l$
Thus number of moles of $N a O H$ required for complete reaction to occur $= 2 \times 0.1 m o l = 0.2 m o l$
3) Thus $0.2 m o l N a O H$ is required for complete reaction.
Since molarity of $N a O H$ is given to be $1 m o l / L$
This means volume of $1 m o l N a O H = 1 L$
Thus volume of $0.2 m o l N a O H = 1 \times 0.2 L = 0.2 L$
$= 0.2 \times 1000 m L = 200 m L$
4) Therefore $200 m L$ of $1 M N a O H$ should be added to $12 g N a H_{2} P O_{4}$ to exactly convert it into $N a_{3} P O_{4}$ .

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$H_{3} P O_{4}$ is a tri basic acid and one of its salt is $N a H_{2} P O_{4}$ . What volume in ml of $1$ M NaOH solution should be added to $12$ g of $N a H_{2} P O_{4}$ to convert it into $N a_{3} P O_{4}$ ?