Number 6.29 ones&Laverman; 59 571572 573 574 575 576 577 578 579 580 5 OUS PAGE
ID: 227856 • Letter: N
Question
Number 6.29 ones&Laverman; 59 571572 573 574 575 576 577 578 579 580 5 OUS PAGE 6.28 A sample of 0.150 M Na,CO,(aq) of volume 25.0 mL is titrated with 0.100 M HCl(aq). What is the pH of the solution at each stoichiometric point in the titration? 6.29 In a "precipitation titration," the concentration of an ion is a with 0.110 M AgNO,(aq). Before the stoichiometric point, the Ag Th fro measured as it forms a precipitate. The concentration of CO,2H ions in a sample of volume 25.0 mL was determined by titrating ions react immediately with the CO ions, but after the stoichio- metric point the concentration of Ag ion increases rapidly. (a) The stoichiometric point is reached after the addition of 36.2 ml of the AgNO,(aq). What is the concentration of CO,2- ions in the sample? (b) The concentration of Ag ions is followed by using a special electrode and measuring pAg (that is,-log [Ag"D. Sketch the plot of pAg against volume of AgNO,(ah and determine the le of H pres that blood value of pAg at the stoichiometric point. 6.30 What volume (in liters) of a saturated mercury(II) sulfide, Ho HgS, so 6.40 consiS lution contains an average of one mercury(II) ion, Hg? 6.31 Two friends go to an all-you-can eat restaurant but eat too much and get heartburn. Both return to their rooms and look for During (a) w a remedy. One friend takes two tablets, each containing 750 mg (rapid a Caro. and the serond friend takes 3 teacmoons of milk of mao about us c areers privacy policy terms of use contactExplanation / Answer
6.29
2AgNO3(aq) + CO3^2-(aq) ---> Ag2CO3(s)
moles AgNO3 used = 0.110 M x 36.2 ml
= 3.982 mmol
moles CO3^2- present = 3.982/2
= 1.991 mmol
molar concentration of Ag2CO3 at stoichiometric point = 1.991 mmol/25 ml
= 0.08 M
Ksp Ag2CO3 = [Ag+]^2.[CO3^2-] = 8.1 x 10^-12
let x amount of Ag2CO3 formed has dissolved
Ag2CO3(s) <==> 2Ag+ + CO3^2-
I 0.08 - -
C - +2x +x
E - 2x x
So,
8.1 x 10^-12 = (2x)^2.(x)
x = 1.26 x 10^-4 M
pAg+ = -log[Ag+]
= -log(2.53 x 10^-4)
= 3.60
so, pAg at stocihiometric point is 3.60
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