Chemistry, 23.09.2019 11:00 mia3128

If solutions of nh4cl(aq) and nh3(aq) are mixed, which ions in the resulting solution are spectator ions in any acid-base chemistry occurring in the solution?

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$\boxed{{\text{C}}{{\text{l}}^ - }}$ ionsact as spectator ions if solutions of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ are mixed.

Further explanation:

Buffer solution offers resistance to any changes in their pH if strong acid or base is added to them. These are usually aqueous solution consisting of weak acid and its conjugate base.

Buffers can be acidic or basic as described below.

Acidic buffer:

These solutions have pH less than 7 and are formed by weak acid and its conjugate base.

Basic buffer:

These solutions have pH more than 7 and are formed by weak base and its conjugate acid.

Spectator ions are those ions that are present in same form on both sides of the chemical reaction. In other words, these ions do not participate in chemical reactions and cannot affect the reaction equilibrium. Since these are not involved in overall chemical reaction, spectator ions are not included while writing net ionic equations for any chemical reaction.

If solutions of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ are mixed, it results in the formation of buffer ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ . is a salt of weak acid and its dissociation occurs as follows:

${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}} \rightleftharpoons {\text{NH}}_4^ + + {\text{C}}{{\text{l}}^ - }$

The reaction of given buffer can be written as follows:

${\text{N}}{{\text{H}}_3} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{NH}}_4^ + + {\text{O}}{{\text{H}}^ - }$

Since ${\text{C}}{{\text{l}}^ - }$ ions are not involved in net chemical reaction for buffer solution of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$ and ${\text{N}}{{\text{H}}_{\text{3}}}$ , these ions act as spectator ions.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Buffer solutions

Keywords: buffer solution, NH4Cl, NH3, NH4+, Cl-, OH-, H2O, buffer, spectator ions.

Answer from: Quest

i think a, hope it .

explanation:

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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If solutions of nh4cl(aq) and nh3(aq) are mixed, which ions in the resulting solution are spectator...

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