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serg [7]
3 years ago
14

A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the

ammonia?
Chemistry
1 answer:
Sveta_85 [38]3 years ago
6 0

Answer:

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

Explanation:

Using the formula

Ca Va = Cb Vb

Cb = 0.32 M

Vb = 50 mL = 50/1000 = 0.050L

Ca = 0.5 M

Va =?

Substituting for Va in the equation, we obtain:

Va = Cb Vb / Ca

Va = 0.32 * 0.05 / 0.5

Va = 0.016 / 0.5

Va = 0.032 L

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

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<u>For (a):</u> The balanced net ionic equation is 2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s) and the sum of coefficients is 4

<u>For (b):</u> The balanced net ionic equation is Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s) and the sum of coefficients is 4

<u>For (c):</u> The balanced net ionic equation is Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s) and the sum of coefficients is

<u>For (d):</u> The balanced net ionic equation is Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s) and the sum of coefficients is 4

<u>For (e):</u> The balanced net ionic equation is Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s) and the sum of coefficients is 3

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Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

  • For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)

The complete ionic equation follows:

2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)

Sum of the coefficients = [2 + 1 + 1] = 4

  • For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)

The complete ionic equation follows:

2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)

Sum of the coefficients = [2 + 1 + 1] = 4

  • For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)

The complete ionic equation follows:

2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)

Sum of the coefficients = [1 + 1 + 1] = 3

  • For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)

The complete ionic equation follows:

2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)

Sum of the coefficients = [2 + 1 + 1] = 4

  • For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)

The complete ionic equation follows:

Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)

Sum of the coefficients = [1 + 1 + 1] = 3

8 0
3 years ago
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