Q.3.105P - Chemistry: Molecular Nature Of Matter And Change

Question

How many grams of NaH₂PO₄ are needed to react with 43.74 mL of 0.285 M NaOH?

$\mathit{N}\mathit{a}{\mathit{H}}_{\mathbf{2}}\mathit{P}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{N}\mathit{a}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{N}{\mathit{a}}_{\mathbf{3}}\mathit{P}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}$

Step-by-Step Solution

Answer

The mass of NaH₂PO₄ needed to react with 43.74 mL of 0.285 M NaOH is 0.747 g of NaH₂PO₄.

Step 1: Calculating the number of moles of HCl

Calculate the number of moles of HCl needed to react with 43.74 mL of sodium hydroxide.

$\begin{array}{rcl}Moles of Na{H}_{2}P{O}_4& =& 43.74\times {10}^{-3}L so\ln \times \frac{0.285 mol HCl}{1 Lso\ln }\times \frac{1 mol Na{H}_{2}P{O}_4}{2 mol Na{H}_{2}P{O}_4}\\ & =& 6.23\times {10}^{-3}mol Na{H}_{2}P{O}_4.\end{array}$

Step 2: Calculating the number of moles

Calculate the mass of NaH₂PO₄as shown below:

$\begin{array}{rcl}Mass of Na{H}_{2}P{O}_4& =& 6.23\times {10}^{-3} mol Na{H}_{2}P{O}_4\times \frac{119.98 g Na{H}_{2}P{O}_4}{1 mol Na{H}_{2}P{O}_4}\\ & =& 0.747 g Na{H}_{2}P{O}_4.\end{array}$