# Question 06667

Jun 23, 2015

$A g C l + 2 N {H}_{4} O H \to \left[A g {\left(N {H}_{3}\right)}_{2}\right] C l + 2 {H}_{2} O$

#### Explanation:

The trick here is to realize that ammonium hydroxide, $N {H}_{4} O H$, is actually ammonia, $N {H}_{3}$, dissolved in water.

$N {H}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

If you write the complete ionic equation for this reaction, you'll get

$A g C {l}_{\left(s\right)} + N {H}_{3 \left(a q\right)} + \cancel{{H}_{2} {O}_{\left(l\right)}} \to \left[A g {\left(N {H}_{3}\right)}_{2}\right] C l + \cancel{{H}_{2} {O}_{\left(l\right)}}$

The net ionic equation for this reaction is

AgCl_((s)) + NH_(3(aq)) -> overbrace([Ag(NH_3)_(2(aq))]^(+))^(color(blue)("diamminesilver (I))) + Cl_text((aq])^(-)#

The ammonia reacts with the silver cations to form the diamminesilver (I) metal complex. To balance this equation, simply multiply the ammonia molecules present on the reactants' side by 2.

$A g C {l}_{\left(s\right)} + 2 N {H}_{3 \left(a q\right)} \to {\left[A g {\left(N {H}_{3}\right)}_{2 \left(a q\right)}\right]}^{+} + C {l}_{\textrm{\left(a q\right]}}^{-}$

To get the initial equation balanced, simply multiply $N {H}_{4} O H$ by 2.

$A g C l + 2 N {H}_{4} O H \to \left[A g {\left(N {H}_{3}\right)}_{2}\right] C l + {H}_{2} O$

Finally, balance the hydrogen and oxygen atoms by multiplying the water molecule present on the products' side by 2.

$A g C l + 2 N {H}_{4} O H \to \left[A g {\left(N {H}_{3}\right)}_{2}\right] C l + 2 {H}_{2} O$

And that's how the balanced chemical equation looks like.