Answer to Question #125161 in General Chemistry for mike

Question

Charlotte was asked to determine the concentration of ammonia in a commercially available cloudy ammonia solution used for cleaning. First she pipetted 25.00mL of the cloudy ammonia solution into a 250.00mL conical flask. 50.00mL of 0.100M HCl(aq) was added to the conical flask which immediately reacted with the ammonia in solution. The excess, (unreacted) HCl was then titrated with 0.050M Na2CO3(aq). 21.50mL of Na2CO3(aq) was required. Help Charlotte out by calculating the concentration of the ammonia in the cloudy ammonia solution.

Answer 1

2HCl_(aq)+Na₂CO_3(aq)→2NaCl_(aq)+CO_2(g)+H₂O_(l)

acid + carbonate→ salt + carbondioxide + water

moles = concentration (mol L^-1) × volume (L)

n(Na₂CO_3(aq)) = c × V

c(Na₂CO_3(aq)) = 0.050 mol L^-1

V(Na₂CO_3(aq)) = 21.50 mL = 21.50 × 10^-3 L

n(Na₂CO_3(aq)) = 0.050 × 21.50 × 10^-3 = 1.075 × 10^-3 mol

From the balanced chemical equation, 1 mole Na₂CO₃ react with 2 moles of HCl

So, 1.075 × 10^-3 mole Na₂CO₃ reacted with 2 × 1.075 × 10^-3 moles HCl

n(HCl_titrated) = 2 × 1.075 x 10^-3 = 2.150 × 10^-3 mol

The amount of HCl that was added to the cloudy ammonia solution in excess was :

n(HCl_exccess) = 2.150 × 10^-3 mol

n(HCl_{total added}) = c × V

c(HCl_{total added}) = 0.100 mol L^-1

V(HCl_{total added}) = 50.00 mL = 50.00 × 10^-3 L

n(HCl_{total added}) = 0.100 × 50.00 × 10^-3 = 5.00 × 10^-3 mol

n(HCl_titrated) + n(HCl_{reacted with ammonia}) = n(HCl_{total added})

n(HCl_{total added}) = 5.00 × 10^-3 mol

n(HCl_titrated) = 2.150 × 10^-3 mol

2.150 × 10^-3 + n(HCl_{reacted with ammonia}) = 5.00 × 10^-3

n(HCl_{reacted with ammonia}) = 5.00 × 10^-3 - 2.150 × 10^-3 = 2.85 × 10^-3 mol

NH_3(aq) + HCl_(aq) → NH₄Cl_(aq)

From the equation, 1 mol HCl reacts with 1 mol NH₃

So, 2.85 × 10^-3 mol HCl had reacted with 2.85 × 10^-3 mol NH₃ in the cloudy ammonia solution

concentration (mol L^-1) = moles ÷ volume (L)

c(NH_3(aq)) = n(NH_3(aq)) ÷ V(NH_3(aq))

n(NH_3(aq)) = 2.85 × 10^-3 mol (moles of NH₃ that reacted with HCl)

V(NH_3(aq)) = 25.00 mL = 25.00 × 10^-3 L (volume of ammonia solution that reacted with HCl)

c(NH_3(aq)) = 2.85 × 10^-3 ÷ 25.00 × 10^-3 = 0.114 mol L^-1