Question

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. (a) 0.260 mole of Ca(NO3)2 in 100.0 mL of solution Ca2+ M? NO3− ?M (b) 5.5 moles of Na2SO4 in 1.25 L of solution Na+? M SO42− ? M (c) 5.00 g of NH4Cl in 490.0 mL of solution NH4+? M Cl − ?M (d) 1.00 g K3PO4 in 300.0 mL of solution K + ? M PO43− ? M

Answer 1

a) Consider dissociation of Ca(NO₃) ₂ in water.

Ca(NO₃) ₂ ( aq) Ca ²⁺ (aq) + 2 NO ₃^- (aq)

From reaction , 1 mol Ca(NO₃) ₂ 1 mol Ca ²⁺ 2 mol NO ₃^-

[ Ca(NO₃) ₂ ] = [ Ca ²⁺ ] and [ NO ₃^- ] = 2 [ Ca(NO₃) ₂ ]

We know that, [ Ca(NO₃) ₂ ] = No. of moles of Ca(NO₃) ₂ / volume of solution in L

[ Ca(NO₃) ₂ ] = 0.260 mol / 0.100 L = 2.60 M

We have, [ Ca(NO₃) ₂ ] = [ Ca ²⁺ ] , therefore, [ Ca ²⁺ ] = 2.60 M

We have, [ NO ₃^- ] = 2 [ Ca(NO₃) ₂ ] , therefore, [ NO ₃^- ] = 2 ( 2.60 M ) = 5.20 M

ANSWER : [ Ca ²⁺ ] = 2.60 M , [ NO ₃^- ] = 5.20 M

b)

Consider dissociation of Na₂SO ₄ in water.

Na₂SO ₄ ( aq) 2 Na ⁺ (aq) + SO ₄^2- (aq)

From reaction , 1 mol Na₂SO ₄ 2 mol Na ⁺ 1 mol SO ₄^2-

[ Na ⁺ ] = 2 [ Na₂SO ₄ ] and [ SO ₄^2- ] = [ Na₂SO ₄ ]

We know that, [ Na₂SO ₄ ] = No. of moles of Na₂SO ₄ / volume of solution in L

[ Na₂SO ₄​​​​​​​ ] = 5.5 mol / 1.25 L = 4.4 M

We have, [ Na ⁺ ] = 2 [ Na₂SO ₄ ] , therefore, [ Na ⁺ ] = 2 ( 4.4 M ) = 8.8 M

We have, [ SO ₄^2- ] = [ Na₂SO ₄ ] , therefore [ SO ₄^2- ] = 4.4 M

ANSWER : [ Na ⁺ ] =  8.8 M , [ SO ₄^2- ] = 4.4 M

C)

Consider dissociation of NH₄Clin water.

NH₄Cl ( aq) NH₄⁺ (aq) + Cl ^- (aq)

From reaction , 1 mol NH₄Cl   1 mol NH₄⁺ 1 mol Cl ^-

[ NH₄Cl ] = [ NH₄⁺ ] = [ Cl ^- ]

We know that, [ NH₄Cl ] = No. of moles of NH₄Cl / volume of solution in L

We have , no. of moles = Mass / Molar mass

No. of moles of NH₄Cl = 5.00 g / 53.49 g/mol = 0.09347 mol

[ NH₄Cl ] = 0.09348 mol / 0.4900 L = 0.191 M

ANSWER : [ NH₄⁺ ] = [ Cl ^- ] = 0.191 M

D )

Consider dissociation of K₃PO₄ in water.

K₃PO₄ ( aq) 3 K ⁺ (aq) + PO₄ ^{3 -} (aq)

From reaction , 1 mol K₃PO₄ 3 mol K ⁺ 1 mol PO₄ ^{3 -}

[ K ⁺ ] = 3 [K₃PO₄ ] and [ PO₄ ^{3 -} ] = [ K₃PO₄ ]

We know that, [ K₃PO₄ ] = No. of moles of K₃PO₄ / volume of solution in L

We have , no. of moles = Mass / Molar mass

No. of moles of K₃PO₄= 1.00 g / 212.27 g/mol = 0.004711 mol

[ K₃PO₄ ] = 0.004711 mol / 0.3000 L = 0.0157 M

We have,  [ K ⁺ ] = 3 [K₃PO₄ ] , therefore  [ K ⁺ ] = 3 ( 0.0157 M ) = 0.0471 M

We have, [ PO₄ ^{3 -} ] = [ K₃PO₄ ] , therefore [ PO₄ ^{3 -} ] = 0.0157 M

ANSWER : [ K ⁺ ] = 0.0471 M , [ PO₄ ^{3 -} ] = 0.0157 M