Enough solid magnesium fluoride is added to water to make a saturated solution.
ID: 999507 • Letter: E
Question
Enough solid magnesium fluoride is added to water to make a saturated solution.
A) Find the concentration of fluoride ion in this solution.
B) Find the concentration of fluoride if the magnesium fluoride is in equilibrium with a 0.00323M solution of magnesium acetate
Pictures below is data if needed to help solve for the problem. Thanks.
1. Dissociation Constants for Acids at 25 °C Formula K. 4.9 × 10-10 3.5x 104 Name Acetic Acetylsalicylic HC9H704 3.3 × 10-4 Adipic Arsenic Arsenous K. K. Name Hydrocyanic HCN Hydrofluoric HF Formula K. K. HC2H302 1.8x 10 H2GH804 3.ax 10-5 3.9 × 10-b HAS04 5.5x 10-3 1.7× 10-7 HaAs03 5.1×10-10 H2C6Hg06 8.0x 10-5 1.6x 1012 HC-H502 6.5 × 10-5 5.1 × 10-12 chromate ion HCrOi 3.0x 107 peroxide 2.4x1012 Benzoic Boric Butanoic Carbonic Chloroacetic Chlorous Citric selenate ion HSe04- Hydrosulfuric H2S Hydrotelluric H2Te Hypobromous HBrO Hypochlorous HCIo HypoiodousHIO lodic Lactic Maleic Malonic 2.2 X 10-2 8.9×10-8 2.3x1023 2.8X 10-9 2.9 x10-8 2.3X10 11 1.7× 10 5.4 × 10-10 1.5×10-5 4.3 ×10-7 1×10 HCaH702 1.6x10-11 5.6 × 10-11 HC-H20:01.4×10-3 HCI02 HaGH507 HCNO HCH02 1.1 × 10 7.4x 10-4 2 × 10 1.8 × 10 2.5 X 10-5 1.7 × 10-5 4.0 × 10-7 HC3H503 1.4x104 Formic H2C4H204 1.2x10-2 5.9x 107 H2C3H204 1.5x10-3 2.0x 10-6 Hydrazoic AppendixI Useful Data A-11 Formula HNO2 H2C20, K. K. K, 4.6 X 10-4 6.0 × 10-2 2.8 × 10-2 5.3 × 10-9 1.3 × 10-10 7.5X103 6.2 x 108 4.2x 1013 5x102 1.3×10-5 4.1×10-3 1.2x 10-1 K, Name Selenous Succinic Sulfuric Sulfurous lartaric Trichloroacetic Trifluoroacetic Name Formula Nitrous ASeQ H2Se03 2.4x103 4.8x109 2.4 × 10-3 6.1×10-5 Paraperiodic H5l06 Phenol Phosphorid Phosphorous H3PO3 Propanoic H2C4H40, 6.2×10-5 2.3×10-6 H2S04Strong acid 1.2x 102 1.6 × 10-2 1.0 × 10-3 2.2 × 10 6.4 × 10-8 4.6×10-5 HGH50 H2C4H406 HC-C1302 2.0x 10 HC3H502 HC3H303 HC-F30, 3.0×10 Pyrophosphoric H4P2O, 7.9 × 10-3 2.0×10-7Explanation / Answer
MgF2 <---> Mg^2+(aq) + 2F^-(aq)
[Mg^2+] = x & [2F^-] = 2x
(x)(2x^2) = 5.6*10^-11
4x^3 = 5.6*10^-11
x^3 = 1.4*10^-11
x = 2.4*10^-4
[Mg^2+] = 2.4*10^-4
2x = [F^-] = 2*(2.4*10^-4 )=4.8*10^-4 M
Find the concentration of fluoride if the magnesium fluoride is in equilibrium with a 0.00323M solution of magnesium acetate
magnesium acetate -- > Mg2++ 2 CH3COO-
0.00323M 0.00323M
MgF2 <---> Mg^2+(aq) + 2F^-(aq)
(0.00323M +X) )(2x^2) = 5.6*10^-11
Here X is very small so we can write 0.00323M +X=0.00323M
(0.00323M )(2x^2) = 5.6*10^-11
0.01292 x^2= 5.6*10^-11
X^2= 4.33*10^-9
X= 6.58*10^-5 M
2x = [F^-] = 2*(6.58*10^-5)=1.32*10^-4 M
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