Question

Part A: What is the concentration of K^+ in 0.15 M of K2S ?

Part B: If $\rm CaCl_2$ is dissolved in water, what can be said about the concentration of the $\rm Ca^{2+}$ ion?

It has the same concentration as the $\rm Cl^-$ ion. Its concentration is half that of the $\rm Cl^-$ ion. Its concentration is twice that of the $\rm Cl^-$ ion. Its concentration is one-third that of the $\rm Cl^-$ ion.

Part C: A scientist wants to make a solution of tribasic sodium phosphate, $\rm Na_3PO_4$ , for a laboratory experiment. How many grams of $\rm Na_3PO_4$ will be needed to produce 700 of a solution that has a concentration of $\rm Na^+$ ions of 0.600 $\it M$ ?

Express your answer numerically in grams.

Answer 1

Concepts and reason

A balance chemical equation of a reaction is the simple representation of the reaction, which gives the idea of the particles and participates in the reaction. Concentration is the amount of substance present in a given volume of a solution.

Mole is the mass that explains the particles present in the substance.

Fundamentals

A balance chemical equation of a reaction provides the number of particles and participates in a chemical reaction.

The molarity of a solution is the number of moles of the solute present in 1L of the solution.

Mole is the mass that explains the particles present in the substance. A mole can be determined by using given molarity and volume.

${\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{ = }}\,\frac{{{\rm{Given}}\,{\rm{mass}}\,{\rm{in}}\,{\rm{gram}}}}{{{\rm{Molecular}}\,{\rm{mass}}}}\,$

${\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{ = }}\,{\rm{Molarity \times }}\,{\rm{Volume}}\,{\rm{in}}\,{\rm{L}}\,$

(A)

Balanced chemical equation of the reaction is

${{\rm{K}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right)\,\longrightarrow{{}}\,{\rm{2}}{{\rm{K}}^{\rm{ + }}}\left( {{\rm{aq}}} \right)\,{\rm{ + }}\,{{\rm{S}}^{{\rm{2 - }}}}\left( {{\rm{aq}}} \right)$

From the chemical equation, one mole of ${{\rm{K}}_{\rm{2}}}{\rm{S}}$ produces two moles of ${{\rm{K}}^{\rm{ + }}}$

Therefore,

$\begin{array}{c}\\{\rm{Concentration}}\,{\rm{of}}\,{{\rm{K}}^{\rm{ + }}}\,{\rm{ = }}\,{\rm{2 \times }}{\rm{.15}}\\\\{\rm{ = }}\,{\rm{0}}{\rm{.30M}}\\\end{array}$

(B)

Balanced chemical equation,

${\rm{CaC}}{{\rm{l}}_{\rm{2}}}\left( {{\rm{aq}}} \right)\,\longrightarrow{{}}\,{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right)\,{\rm{ + 2C}}{{\rm{l}}^{\rm{ - }}}\,\left( {{\rm{aq}}} \right)$

From the equation,

$\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\,{\rm{ = }}\,\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{C}}{{\rm{l}}^{\rm{ - }}}} \right]$

Therefore, concentration of the ${\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}$ ion is half of the concentration of ${\rm{C}}{{\rm{l}}^{\rm{ - }}}$ ion.

(C)

${\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{ = }}\,{\rm{Molarity \times }}\,{\rm{Volume}}\,{\rm{in}}\,{\rm{L}}\,$

$\begin{array}{c}\\{\rm{moles}}\,{\rm{of}}\,{\rm{N}}{{\rm{a}}^{\rm{ + }}}\,{\rm{ions}}\,{\rm{ = }}\,{\rm{600 \times 700 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\\\\\,{\rm{ = }}\,{\rm{0}}{\rm{.420}}\\\end{array}$

Balanced chemical equation,

${\rm{N}}{{\rm{a}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\left( {{\rm{aq}}} \right)\,\longrightarrow{{}}\,{\rm{3N}}{{\rm{a}}^{\rm{ + }}}\left( {{\rm{aq}}} \right)\,{\rm{ + }}\,{\rm{PO}}_{\rm{4}}^{{\rm{3 - }}}\left( {{\rm{aq}}} \right)$

From the equation, one mole of ${\rm{N}}{{\rm{a}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}$ produces three moles of ${\rm{N}}{{\rm{a}}^{\rm{ + }}}$ ions.

Therefore,

$\begin{array}{c}\\{\rm{moles}}\,{\rm{of}}\,{\rm{N}}{{\rm{a}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\,{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.420}}}}{{\rm{3}}}\\\\{\rm{ = }}\,{\rm{0}}{\rm{.140}}\\\\{\rm{molecular mass of N}}{{\rm{a}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\,{\rm{ = }}\,{\rm{163}}{\rm{.94g}}\\\\{\rm{mass}}\,{\rm{of}}\,{\rm{N}}{{\rm{a}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.140 \times 163}}{\rm{.94}}\\\\{\rm{ = }}\,{\rm{22}}{\rm{.95}}\,{\rm{g}}\\\end{array}$

Ans: Part A

${\rm{Concentration}}\,{\rm{of}}\,{{\rm{K}}^{\rm{ + }}}\,{\rm{ = }}\,{\rm{0}}{\rm{.30M}}$

Part B

Part C

${\rm{mass}}\,{\rm{of}}\,{\rm{N}}{{\rm{a}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{4}}}{\rm{ = }}\,{\rm{22}}{\rm{.95}}\,{\rm{g}}$