NO2 + H 2O 0 100 x in C#

Decoder DataMatrix in C# NO2 + H 2O 0 100 x

NO2 + H 2O 0 100 x
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OH + HNO2 x x
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K b = 10 1065 = 2 24 10 11 =
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( x )(x ) neglect x 0 100 x x = [OH ] = 1 50 10 6 M pOH = 5 82 O pH = 14 00 pOH = 14 00 5 8 2 = 8 18
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e 5500 mL Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem Stoichiometry requires a balanced chemical equation and moles HNO2 + NaOH Na + + NO + H 2O 2 0300 mol Base: 5500 mL = 00165 mol 0 1000 mL Based on the stoichiometry of the problem, and on the moles of acid and base, the acid is now the limiting reagent HNO2 + NaOH Na + + NO + H2O 2 0 0150 00165 mol 0 0 00150 00150 +00150 +00 1 50 00000 0000 00150
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228 U Step 4 Review the Knowledge You Need to Score High
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The strong base will control the pH
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3 [OH ] = 00015 mol/0155 L = 97 10 M
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The stoichiometry part of the problem is finished Since this is now a solution of a strong base, it is now a simple pOH/pH problem
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3 pOH = log 97 10 = 201
pH = 1400 pOH = 1400 201 = 1199 f 7500 mL Since both an acid and a base are present (and they are not conjugates), this must be a stoichiometry problem Stoichiometry requires a balanced chemical equation and moles
HNO2 + NaOH Na + + NO2 + H 2O 0300 mol Base: 7500 ML = 00225 mol 1000 mL
Based on the stoichiometry of the problem, and on the moles of acid and base, the acid is now the limiting reagent HNO2 + NaOH Na + + NO + H2O 2 0 0150 00225 mol 0 0 00150 00150 +00150 +0 0150 00000 000 75 00150
init react t final
The strong base will control the pH
2 [OH ] = 00075 mol/0175 L = 43 10 M
The stoichiometry part of the problem is finished Since this is now a solution of a strong base, it is now a simple pOH/pH problem
2 pOH = log 43 10 = 137
pH = 1400 pOH = 1400 137 = 1263
Solubility Equilibria
Many salts are soluble in water, but some are only slightly soluble These salts, when placed in water, quickly reach their solubility limit and the ions establish an equilibrium system with the undissolved solid For example, PbSO4, when dissolved in water, establishes the following equilibrium: PbSO 4 (s) Pb2+ (aq) + SO2 (aq) 4
The equilibrium constant expression for systems of slightly soluble salts is called the solubility product constant, Ksp It is the product of the ionic concentrations, each one raised to the power of the coefficient in the balanced chemical equation It contains no denominator since the concentration of a solid is, by convention, 1 and does not appear in the equilibrium constant expressions (Some textbooks will say that the concentrations of
Equilibrium 229
solids, liquids, and solvents are included in the equilibrium constant) The Ksp expression for the PbSO4 system would be: K sp =[Pb2+ ][ SO2 ] 4 For this particular salt the numerical value of Ksp is 16 10 8 at 25 C Note that the 2+ 2 Pb and SO4 ions are formed in equal amounts, so the right-hand side of the equation 2 could be represented as [x] If the numerical value of the solubility product constant is known, then the concentration of the ions can be determined And if one of the ion concentrations can be determined, then Ksp can be calculated 8 For example, the Ksp of magnesium fluoride in water is 8 10 How many grams of magnesium fluoride will dissolve in 0250 L of water MgF2 ( s) Mg 2+ + 2F K sp = [ Mg 2+ ][F ]2 = 8 10 8 = ( x )( 2x )2 = 4 x 3 = 8 10 8 x = 3 10 3 = [ Mg 2+ ] (1 mol MgF2 )(62 3g MgF2 ) (3 10 3 mol Mg 2+ ) (0 250L) 0 (L ) (1 mol Mg 2+ )(1 mol MgF2 )
= 0 05 g If a slightly soluble salt solution is at equilibrium and a solution containing one of the ions involved in the equilibrium is added, the solubility of the slightly soluble salt is decreased For example, let s again consider the PbSO4 equilibrium: PbSO 4 (s) Pb2+ ( aq) + SO2 ( aq) 4
Suppose a solution of Na2SO4 is added to this equilibrium system The additional sul^ fate ion will disrupt the equilibrium, by Le Cha telier s principle, and shift it to the left, decreasing the solubility The same would be true if you tried to dissolve PbSO4 in a solution of Na2SO4 instead of pure water the solubility would be lower This application of ^ Le Cha telier s principle to equilibrium systems of slightly soluble salts is called the common-ion effect Calculations like the ones above involving finding concentrations and Ksp can still be done, but the concentration of the additional common ion will have to be inserted into the solubility product constant expression Sometimes, if Ksp is very small and the common ion concentration is large, the concentration of the common ion can simply be approximated by the concentration of the ion added For example, calculate the silver ion concentration in each of the following solutions: a Ag2CrO4(s) + water b Ag2CrO4(s) + 100 MNa2CrO4 Ksp = 19 10 12 a Ag 2CrO 4 (s) 2 Ag + ( ag ) + CrO 4 ( ag )
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