Solubility Questions and Answers

Solubility Solubility Questions and answers Question 1 Show, by calculation, that a precipitate forms when 50.0 mL of 0.100 mol L –1 KClaq  KClaq!! and 50.0 mL  –1 of 0.0"00 mol L  #$%&'aq! aq! are mi(ed. K s #$Cl! ) 1.5* + 10 –10  After mixing volume = 100 100 mL [Ag+ ] = 0.020   2 = 0.01 mol L –1 [Cl ] – ] = 0.100   2 = 0.05 mol L–1 IP = [Ag+ ][Cl– ]  ] = 0.01 ⋅ 0.05 = 5 ⋅ 10–4 IP >  ! !o "re#i"it$te form!.

Question "  # 50.0 50.0 mL sample sample of a saturated aueous aueous solution solution of lead bromide, bromide, -br ", was e/aporated to dryness. 0."" $ of solid -br " was obtained. a! a!

i! i!

rit rite e the the eua euati tion on for for the the eu euil ilib ibri rium um pre prese sent nt in in a sat satur urat ated ed sol solut utio ion n of lea lead d brom bromid ide. e.

ii! ii!

Compl omplet ete e the the e(p e(pre ress ssio ion n for for K s-br "!.

 %$  % $&

(b) (b)

%i&

P'(r2 %!&

2+

 P'   P'

%$)& + 2(r *%$)&

Calcul lculat ate e the the val value of K s(PbBr  ). 2  2 

 – M (PbBr 2   )  = 367 g mol  2 %'& n%P'(r 2 & in 50 mL = 0.422   ,,= 1.15 ⋅ 10 mol n%P'(r 2 & in 1 L = 1.15 ⋅ 10 ⋅ 20 = 2.0 ⋅ 10 mol !olu'ilit%!& of P'(r 2 = 2.0 ⋅ 10 mol L or /ire#tl n%P'(r 2 & in 1 L = .44   ,,= 2.0 ⋅ 10 mol [P'  ] =2.0 ⋅ 10 mol L [(r  ]= 4.,0  10 mol L  ! = [P'  ][(r  ]2 = 2.0 ⋅ 10 ⋅ %4.,0 ⋅ 10  &2 = 4.- ⋅ 10 r 4! = 4%2.0 ⋅ 10  & –

–

–2 –2

–2

2+

–

2+

–2

*1

–2

*1

–

–2

–2

–5



–2 

 != 4.- ⋅ 10 % !.f.& –5

*1

%ii&

! =  [P'  ][(r  ]  ! ex"re!!ion 2+

– 2

Question ' (a) (a)

Desc Descri ribe be what what is mean meantt by by the the term term ‘solubility’.

%$& 3olu'ilit i! te $mount of !u'!t$n#e t$t ill /i!!olve in $ given volume to form $ !$tur$te/ !olution %$t t$t tem"er$ture&.

(b) (b)

The The solu solubi bili lity ty prod produc uct, t,  s, of AgCl has a alue of !."# × !$ %!$ at &"'C and this alue increases to &.!" × !$ % at !$$'C. *plain why  s is higher at !$$'C. +nclude reference to the releant euilibrium euation in your answer.

%'&  ! i! $n e)uili'rium e)uili'rium #on!t$nt for te re$#tion AgCl%!& Ag+ %$)& + Cl– %$)& AgCl%!& 6 Ag+%$)& + Cl–%$)& $n/  ! = [Ag+ ][Cl ] – ] 7ore !oli/ /i!!olve! en te tem"er$ture i! in#re$!e/ $! e)uili'rium !ift! in te en/otermi# /ire#tion8 i# me$n! it !ift! to te rigt $n/ i n#re$!e! te #on#entr$tion of ion! in !olution. 9i! in#re$!e!  !. An!er re#ogni!e! t$t in#re$!e/  ! v$lue me$n! in#re$!e/ !olu'ilit or #on#entr$tion of ion!

The chloride ion concentration in sea water can be determined by titrating a sample with aueous siler nitrate (Ag-/) using potassium chromate (0 &Cr1) as the indicator. As the siler nitrate is added, a precipitate of siler chloride, (AgCl) forms. 2hen most of the AgCl has precipitated, the Ag3($)) concentration becomes high enough for a red precipitate of Ag &Cr1 to form. (c) (c)

4how 4how that that the the sol solub ubil ilit ity y of of Ag Ag&Cr1 in pure water at &" oC is higher than that of AgCl.

 s(AgCl) 5 !."# × !$ %!$

 s(Ag&Cr1) 5 !./$ × !$ %!&

%#& %i&  [Ag+ ] = [Cl– ]  ] = !olu'ilit8 !olu'ilit8 !  ! = ! !o ! =: ! = 1.25 ⋅ 10 mol L 2

i i&

–5

–1

! =  [Ag ] [Cr4  ] + 2

 2–

= ! = ,. ⋅ 10 mol L –5

–1

(d) +f the the concen concentrat tration ion of of chroma chromate te ions ions is #./$ #./$ 6 !$ !$ %/ mol 7 %! at the point when the Ag&Cr1 starts to precipitate, calculate the concentration of Ag3 ions in the solution. (d)