Problem 3

Silver chloride is a milk-white solid (quotes from a lesson by L. J. Gay-Lussac). Data at 298 K:

pK_{s1}(\ce{AgCl}) = 9.7

;

pK_{s2}(\ce{Ag2CrO4}) = 12

; formation constant of

[\ce{Ag(NH3)_n}]^+

\beta_n = 10^{7.2}

;

E^\circ(\ce{Ag+/Ag}) = 0.80

V. (a) Calculate the solubility

s

\ce{AgCl(s)}

in water. (b) When ammonia is added to silver chloride a complex of stoichiometry

n

is formed; write the equilibrium and its constant

K

, then determine

n

knowing that

0.1

mol of

\ce{AgCl}

in 1 dm

^3

of water just dissolves when

[\ce{NH3}] = 1.78

mol dm

^{-3}

. (c) The Mohr method titrates

\ce{Cl-}

\ce{Ag+}

in the presence of

\ce{K2CrO4}

: three drops (

\approx 0.5

^3

) of

\ce{K2CrO4}

c = 7.76\times 10^{-3}

mol dm

^{-3}

are added to

V_0 = 20.00

^3

of a NaCl solution titrated by

\ce{AgNO3}

c = 0.050

mol dm

^{-3}

; a red precipitate appears at

V_{\ce{Ag}} = 4.30

^3

. Calculate

c(\ce{Cl-})

and the residual

[\ce{Cl-}]_{\text{res}}

when

\ce{Ag2CrO4}

starts to precipitate.

Step 3 of 5: Stoichiometry of the complex

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n = \dfrac{\log\big([\ce{Ag(NH3)_n}^+][\ce{Cl-}]/K\big)}{\log[\ce{NH3}]} \Rightarrow n = 2

Analysis

When the last grain dissolves, $[\ce{Ag(NH3)_n}^+] = [\ce{Cl-}] = 0.1$ mol dm $^{-3}$ and $[\ce{NH3}] = 1.78$ mol dm $^{-3}$ . Then $[\ce{NH3}]^n = \dfrac{0.1\times 0.1}{10^{-2.5}} = 3.16$ , so $n = \log 3.16 / \log 1.78 = 2$ : the complex is $[\ce{Ag(NH3)2}]^+$ .