(a) The molar solubility of PbBr_{2}\ at\ 25^{\circ}C\ is\ 1.0 \tim

Tiffany Russell 2021-12-18 Answered

(a) The molar solubility of PbBr2 at 25C is 1.0×102molL. Calculate Kp- (b) If 0.0490 g of AgIO3 dis- sp• solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate Ksp value from Appen- dix D, calculate the pH of a saturated solution of Ca(OH)2.

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Expert Answer

Alex Sheppard
Answered 2021-12-19 Author has 36 answers

Step 1
The molar solubility of PbBr2 is 1.0×102malL. The amount of AgIO3 dissolved per liter of solution is 0.0490 g. The value of Ksp of Ca(OH)2 is 6.5×106.
Step 2
(a) The value of Ksp for compound with general formula AX2 is calculated as,
Ksp=4s3
Where,
- s is the molar solubility of compound.
Substitute the value of molar solubility of PbBr2 in the above formula.
Ksp=4(1.0×102)3
=4.0×106
The solubility product constant of PbBr2 is 4.0×106.
Step 3
(b) The number of moles of AgIO3 is calculated by the formula,
nAgIO3=mM
Where,
- m is the given mass of AgIO3.
- M is the molar mass of AgIO3.
The molar mass of AgIO3 is 282.77gmol.
Substitute the given mass and molecular mass of AgIO3 in the above formula.
nAgIO3=0.0490g282.77gmol
=1.73×104mol
Hence, the molar solubility of AgIO3 is 1.73×104M.
The value of Ksp for compound with general formula AX is calculated as,
Ksp=s2
Substitute the molar solubility of AgIO3 in the above formula.
Ksp=(1.73×104)2
=3.0×108
The solubility product constant of AgIO3 is 3.0×108.
Step 4
(c) The value of Ksp for compound with general formula AX2 is calculated as,
Ksp=4s3
Substitute the value of Ksp for Ca(OH)2 in the above formula.
6.5×106=4s3

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Elois Puryear
Answered 2021-12-20 Author has 30 answers

(a) molar solubility=1.0×102molL
PbBr2Pb2+2Br
[Pb2+]=1.0×102molL
[Br]=2×1.0×102molL=2.0×102molL
Ksp=[Pb2+][Br]2=(1.0×102)(2.0×102)2=4.0×106
enable the molar solubilit be S Then [Ca2+]=S[IO3]2=2S The solubility product =S×(2S)2=4S3=7.a million×107=710×109 which on fixing yields S=8.9×103 and the molar solubility For the subsequent party, you consider x because the molar solubility of Ca(IO3)2 in water and then the most objective of IO3 must be (2x+0.06) for the reason that NaIO3 is carefully soluble and then [Ca2+]=x then writing the solubility product equation x(2x+0.06)=7.a million×107 and be certain for x, it quite is the molar solubility Ca(IO3)2 in the presence of 0.06M NaIO3.
(b) The reaction AgIO3(s)Ag+(aq)+IO3
0.0490g of AgIO3 dissolves per liter of solution.
Let us calculate the solubility-product constant.
First, let us find the molar solubility of AgIO3.
The molar mass of AgIO3 is
MAgIO3=MAg+MI+3MO

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