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Consider the reaction between $\mathrm{Na_2CO_3(aq)}$ and $\mathrm{H_3PO_4(aq)}$ as shown in the following balanced equation: $\newline$ $3\mathrm{Na_2CO_3(aq)} + 2\mathrm{H_3PO_4(aq)} \rightarrow 2\mathrm{Na_3PO_4(aq)} + 3\mathrm{H_2O(l)} + 3\mathrm{CO_2(g)}$ $\newline$ \begin{array}{ll} $\newline$ \text{Chemical} & \text{MW} (\newline\)\mathrm{Na_2CO_3} & $105$ . $991$ (\newline\)\mathrm{H_3PO_4} & $97$ . $9977$ (\newline\)\mathrm{Na_3PO_4} & $163$ . $944$ (\newline\)\mathrm{H_2O} & $18$ . $0158$ (\newline\)\mathrm{CO_2} & $44$ . $011$ (\newline\)\end{array} $\newline$ What is the concentration of the phosphoric acid solution if $199.2\,\mathrm{mL}$ of it generates $345.4\,\mathrm{mL}$ of a $1.23\,\mathrm{M}$ sodium phosphate for complete reaction?

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Ratio and Quadratic equation

Full solution

Q. Consider the reaction between

\mathrm{Na_2CO_3(aq)}

and

\mathrm{H_3PO_4(aq)}

as shown in the following balanced equation:

\newline

3\mathrm{Na_2CO_3(aq)} + 2\mathrm{H_3PO_4(aq)} \rightarrow 2\mathrm{Na_3PO_4(aq)} + 3\mathrm{H_2O(l)} + 3\mathrm{CO_2(g)}

\newline

\begin{array}{ll}

\newline

\text{Chemical} & \text{MW} (\newline\)\mathrm{Na_2CO_3} &

105

.

991

(\newline\)\mathrm{H_3PO_4} &

97

.

9977

(\newline\)\mathrm{Na_3PO_4} &

163

.

944

(\newline\)\mathrm{H_2O} &

18

.

0158

(\newline\)\mathrm{CO_2} &

44

.

011

(\newline\)\end{array}

\newline

What is the concentration of the phosphoric acid solution if

199.2\,\mathrm{mL}

of it generates

345.4\,\mathrm{mL}

of a

1.23\,\mathrm{M}

sodium phosphate for complete reaction?

Determine Moles of Na $_3$ PO $_4$ : First, we need to determine the moles of sodium phosphate produced in the reaction. $\newline$ Moles of Na $_3$ PO $_4$ = Molarity $(M)$ * Volume $(L)$
Convert Volume to Liters: Convert the volume of sodium phosphate from mL to L. Volume of $\text{Na}_3\text{PO}_4$ in liters = $345.4\,\text{mL} \times \left(\frac{1\,\text{L}}{1000\,\text{mL}}\right) = 0.3454\,\text{L}$
Calculate Moles of Na $3$ PO $4$ : Calculate the moles of Na $3$ PO $4$ produced. $\newline$ Moles of Na $3$ PO $4$ = $1.23 \, \text{M} \times 0.3454 \, \text{L} = 0.424842 \, \text{moles}$
Use Stoichiometry for $\text{H}_3\text{PO}_4$ : Using the stoichiometry of the balanced equation, we can find the moles of $\text{H}_3\text{PO}_4$ that reacted. According to the balanced equation, $3$ moles of $\text{Na}_2\text{CO}_3$ react with $2$ moles of $\text{H}_3\text{PO}_4$ to produce $2$ moles of $\text{Na}_3\text{PO}_4$ .
Calculate Moles of H $3$ PO $4$ : Calculate the moles of H $3$ PO $4$ that reacted using the mole ratio from the balanced equation. $\newline$ Moles of H $3$ PO $4$ = $(\frac{2 \text{ moles H3PO4}}{2 \text{ moles Na3PO4}}) \times \text{Moles of Na3PO4}$ $\newline$ Moles of H $3$ PO $4$ = $(\frac{2}{2}) \times 0.424842 \text{ moles} = 0.424842 \text{ moles}$
Find Concentration of H $\text{H}_3\text{PO}_4$ : Now, we need to find the concentration of the H $\text{H}_3\text{PO}_4$ solution. Concentration is moles of solute divided by the volume of solution in liters.
Convert Volume to Liters: Convert the volume of $\text{H}_3\text{PO}_4$ solution from mL to L. $\newline$ Volume of $\text{H}_3\text{PO}_4$ in liters = $199.2\,\text{mL} \times \left(\frac{1\,\text{L}}{1000\,\text{mL}}\right) = 0.1992\,\text{L}$
Calculate Concentration of H $3$ PO $4$ : Calculate the concentration of the H $3$ PO $4$ solution. $\newline$ Concentration of H $3$ PO $4$ = Moles of H $3$ PO $4$ / Volume of H $3$ PO $4$ in liters $\newline$ Concentration of H $3$ PO $4$ = $\frac{0.424842 \text{ moles}}{0.1992 \text{ L}}$
Perform Division: Perform the division to find the concentration. $\newline$ Concentration of $\mathrm{H_3PO_4} = 2.133\,\mathrm{M}$ (rounded to three decimal places)

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