8) The addition of fluoride to many public water systems is done to help fight c

Question

8) The addition of fluoride to many public water systems is done to help fight cavities in a city's population. Given 8) that a 500 mL sample of hard water contains 2.5x 10 MCa ions, what is the maximum amount of sodium fuoride, in grams, that can be dissolved in the water? ') B) 11 x o (Ksp [CaF2] = 5.3 x 10 A) 2.1 x 10 c) 4.5 × 10 D)2.5×10 E)5.3 x Io 9) 9) If solid NaOH is added to an aqueous solution of 0.050 M Fe(NO3)2. at what pH will Fe(O2() begin to precipitate? K Felo2 4 10 5 = 4.1 x 10 A) 7.46 B) 654 C) 10.46 D) 12.82 E) 8.38 10) 10) What will be the effect on the following system if a catalyst is added? Ca2+(gg) C2042. (gg) CaC204 (s) + A) precipitation of Ca B) precipitationof CaC204 C) Ks increases D) an increase in [Ca E) There is no change in the equilibrium 11) If SSS mg of NaC2H302 (FW 82.04) is added to 250. mL of 0.015 M P(NO32. will a precipitate form? (Ken lead acetate = 1.8 10 .) A) No, because Qin> Ksp B) No, because Qip s C) Yes, because Qip Ksp D) No, because QipKsp E) Yes, because Qip "Ksp 12)Two salts. AX and BX2.have identical Ksp values at a given temperature. We can say 12) A) the molar solubility of BX2 is greater than that of AX B) addition of NaX will not affect the solubilities of the salts. C) the salts are more soluble in 0.1 MNaX than in water. D) the molar solubility of AX is greater than that of BX2 E) the molar solubility of AX is identical to BX2

Explanation / Answer

Q12

D

If they have the identical Ksp then we can say that Ksp of AX=S2 and Ksp =4*S13 hence if they have the same Ksp then S>S1

Q11

Moles of Acetate 0.555/82.04=0.00676
Conc of Acetate 0.00676/0.25=0.027 M
Conc of Lead=0.015

Q of Lead Acetate=0.015*(0.027)2 =1.09*10-5 < Ksp Hence does not precipitate

Q10

There is no effect of adding a catalyst in equilibrium.

Q9

[OH-]2=Ksp/[Fe]=82*10-15

Hence pOH=-log([OH-]2)/2=6.543

Hence pH=14-pOH=7.46

Q8

I am getting the answer 3*10-2

According to the calculations

[Ca][F]2=5.3*10^-9

[F]2=(5.3*10^-9)/2.5*10^-3 Hence [F]=1.45*10^-3

Hence Mass of NaF=1.45*10^-3*0.5*42=0.03

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