Neutral solutions in terms of pH, pOH, moles of acid and bases
For our problems the idea of neutral solutions has meaning only for water solutions. Neutral solutions are defined in two ways. One definition is in terms of pH. The other is in terms of concentrations of hydronium ion and hydroxide ion. |
Neutral solutions pH, pOH, pKw One definition of neutrality is in terms of pH. At 25oC neutral solutions have pH = 7. This comes from two sorces. One is the value for the ion product for wate. The second is the definition of the operator "p" or "-log". The symbol "p" means carry out the operation of taking the negative log of any number that follows the symbol. If we apply this "operator" to the number value for the ion product for water the following equations are created. EXAMPLE: What is the "p" for 100 or 102? ANSWER: -log[1 x 102 ] = -2 EXAMPLE:
The equation means that a solution is neutral when the acid potential and base potential are equal. The pH must equal the pOH. At 25oC this means pH = pOH = 7
A very important idea is buried in these equations. The ion product for water, Kw, is 1 x 10-14 only at 25oC. The self ionization of water ( the dissociation of water) is endothermic. This means dissociation increase when the temperature increase. The ion product gets bigger and changes with temperature. The value for Kw is bigger at higher temperatures. For example at human body temperature which is 37oC the value for the ion product is Kw =2.4 x 10-14. This means water solutions in the body are neutral when pH = 6.81.
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Concentrations of hydronium, [H3O1+], and hydroxide ion, [OH1-]
A neutral solution is also defined in terms of the numbers of the concentrations of hydronium ion and hydroxide ion. When the two concentrations are equal the solution is neutral. This provides one basis for neutralization calculations.
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Moles of hydronium, [H3O1+], and mole of hydroxide ion, [OH1-] A neutral solution is also defined in terms of the numbers of moles of hydronium ion and moles of hydroxide ion. When the mole amounts are equal the solution is neutral. This provides another basis for neutralization calculations.
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Example: How many milliliters of 0.45 M HCl solution are needed to neutralize 20.0 mL of 0.15 M KOH. The fundamental relationship to use here is MaVa = MbVb This question is answered by solving for the volume of acid, Va = (MbVb ) / (Ma) Replace symbols with the values from the data. Va = (0.15 M KOH x 0.020 Liter) / (0.45 M HCl) Va = (0.0030 A volume of 6.7 mL 0.45 M HCl is needed.
MaVa = MbVb (0.45 M HCl) Va = (0.15 M KOH x 20 milliliter)Va= (0.15 M KOH x 20 milliliter) / (0.45 M HCl) Va = (30 millimoles) / ( 0.45 moles/
liter) = (30 milli Va = (30 milli |