(a) The molar solubility of PbBr2 at 25∘C is 1.0×10−2molL. 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.

Question

(a) The molar solubility of PbBr2 at 25∘C is 1.0×10−2molL. 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.

Alex Sheppard · Accepted Answer

Step 1 The molar solubility of PbBr2 is 1.0×10−2malL. The amount of AgIO3 dissolved per liter of solution is 0.0490 g. The value of Ksp of Ca(OH)2 is 6.5×10−6. 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(1.0×10−2)3 =4.0×10−6 The solubility product constant of PbBr2 is 4.0×10−6. 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 =1.73×10−4mol Hence, the molar solubility of AgIO3 is 1.73×10−4M. 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=(1.73×10−4)2 =3.0×10−8 The solubility product constant of AgIO3 is 3.0×10−8. 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. 6.5×10−6=4s3

Elois Puryear · Accepted Answer

(a) molar solubility=1.0×10−2molL PbBr2⇒Pb2+2Br− [Pb2+]=1.0×10−2molL [Br−]=2×1.0×10−2molL=2.0×10−2molL Ksp=[Pb2+][Br−]2=(1.0×10−2)(2.0×10−2)2=4.0×10−6 enable the molar solubilit be S Then [Ca2+]=S[IO3]2−=2S The solubility product =S×(2S)2=4⋅S3=7.a million×10−7=710×10−9 which on fixing yields S=8.9×10−3 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×10−7 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+3⋅MO

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

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