If 4.15 L of a 0.800 M H3PO4 solution is diluted to a volume of 10.00 L, what is the...

Given Data:

• The volume of {eq}{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} solution before dilution: {eq}{{\rm{V}}_1} = 4.15\;{\rm{L}} {/eq}.

• The concentration of {eq}{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} solution before dilution: {eq}{{\rm{M}}_1} = 0.800\;{\rm{M}} {/eq}.

• The volume of {eq}{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} solution after dilution: {eq}{{\rm{V}}_2} = 10.00\;{\rm{L}} {/eq}.

To Calculate:

• The concentration of {eq}{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} {/eq} after dilution that is {eq}{{\rm{M}}_2} {/eq}.

The formula is given below.

{eq}{{\rm{M}}_1}{{\rm{V}}_1} = {{\rm{M}}_2}{{\rm{V}}_2} {/eq}

Where,

• {eq}{{\rm{M}}_1}\;{\rm{and}}\;{{\rm{M}}_2} {/eq} are the concentration before and after dilution respectively.

• {eq}{{\rm{V}}_1}\;{\rm{and}}\;{{\rm{V}}_2} {/eq} are the volume before and after dilution respectively.

On substituting the corresponding values in the above equation, we get,

{eq}\begin{align*} 0.800\;{\rm{M}} \times 4.15\;{\rm{L}} &= {{\rm{M}}_2} \times 10.00\;{\rm{L}}\\ {{\rm{M}}_2} &= \dfrac{{0.800\;{\rm{M}} \times 4.15\;{\rm{L}}}}{{10.00\;{\rm{L}}}}\\ &= 0.332\;{\rm{M}} \end{align*} {/eq}

Hence, the concentration of the resulting solution is 0.332 M.