Question

12.) determine the number of ml of 0.246 m agno₃ required to precipitate all the phosphate ion in a solution containing 2.10 g of sodium phosphate. hint: write the balanced chemical equation first. include the correct number of significant figures and the correct unit. 156.3 ml

Explanation:

Step1: Calculate moles of sodium phosphate ($\ce{Na3PO4}$)

The molar mass of $\ce{Na3PO4}$ is $M = 3\times22.99+30.97 + 4\times16.00=163.94\ g/mol$.
Using the formula $n=\frac{m}{M}$, where $m = 2.10\ g$ and $M = 163.94\ g/mol$.
$n(\ce{Na3PO4})=\frac{2.10\ g}{163.94\ g/mol}\approx0.0128\ mol$.

Step2: Use stoichiometry from balanced equation

The balanced chemical equation is $3\ce{AgNO3}+\ce{Na3PO4}=\ce{Ag3PO4}\downarrow+3\ce{NaNO3}$.
From the equation, the mole ratio of $\ce{AgNO3}$ to $\ce{Na3PO4}$ is $3:1$.
So, $n(\ce{AgNO3}) = 3\times n(\ce{Na3PO4})=3\times0.0128\ mol = 0.0384\ mol$.

Step3: Calculate volume of $\ce{AgNO3}$ solution

Using the formula $V=\frac{n}{C}$, where $n = 0.0384\ mol$ and $C=0.246\ M$.
$V=\frac{0.0384\ mol}{0.246\ mol/L}\approx0.156\ L$.
Convert liters to milliliters: $V = 0.156\ L\times1000\ mL/L = 156\ mL$.

Answer:

$156\ mL$