In the English language, changing the suffix of a word like baked to baking might make it conform to the rules of grammar, but it won’t dramatically change the meaning of the word. That’s not the case when it comes to the language of chemistry.
Consider aluminum phosphide versus aluminum phosphate. Aluminum phosphide (AlP) is a common and inexpensive pesticide. But it releases deadly phosphine gas when it comes in contact either with moisture or with hydrochloric acid in the stomach , as shown by the following reactions:
AlP + 3 H2O → Al(OH)3 + PH3↑
AlP + 3 HCl → AlCl3 + PH3↑
Aluminum phosphide-poisoning is a serious problem in the Middle East and Asia. It has also led to rare but profound tragedies in the United States. Four children were killed in Texas last year (2017) after Fumitoxin pellets near their mobile home were hosed down by their father after neighbors complained about the smell.
Change phosphide’s suffix from ide to ate, and you get a totally different compound known as aluminum phosphate (AlPO4). This substance is formed from the reaction between aluminum sulfate and phosphate in waste water. Since AlPO4 is poorly soluble, the phosphate is unavailable to algae and eutrophication becomes less likely.
The maximum solubility of aluminum phosphate at a given temperature can be calculated from its equilibrium solubility product constant, known as Ksp.
For AlPO4 the expression is given by:
Ksp= [ Al 3+][ PO43-],
where the square brackets represent the equilibrium concentration of the respective ions in moles/L. Since the stoichiometric ratio of AlPO4 to Al 3+ is one to one, the latter’s concentration is equal to how many moles of AlPO4 dissolved per liter of solution.
But what if there was already some PO43- present in a solution before AlPO4 appeared? In the same manner that any phosphate naturally present in an individual’s mouth makes his teeth enamel less vulnerable to bacterial acidic attack, phosphate’s present will decrease the solubility of AlPO4 .
Let’s say there are 0.050 M of phosphate already present, then the Ksp expression becomes:
Ksp= [ x ][ x +0.050] . Since Ksp for AlPO4 is 9.84 X 10-21. Expanding we get:
x2 + 0.050x – 9.84 X 10-21 = 0.
But if this is solved with a typical calculator, it will yield an answer of zero. this happens because in the quadratic formula based on ax2+ bx + c = 0 :
b2 >> 4ac, which means that the square rooted expression yields b to the typical calculator. ± b, when combined with –b from the formula’s numerator yields either the useless negative value and zero.
What’s the solution then?
In such cases we can use an alternate version of the quadratic formula, which can be obtained by multiplying the numerator and denominator by –b ±√( b2– 4ac). This will yield:
With this version of the formula, when b2 >> 4ac, we get either an undefined expression (division by zero) which we ignore, or approximately c/-b, which in our example gives us the solubility(x) to be not quite zero (although awfully close!) but 1.97 X 10-19 mol/L. More generally this version increases accuracy anytime the absolute value of one of the roots is much smaller than the other. A calculator can be easily programmed with conditional statements so that it can handle any situation. I wrote this one for a TI-83.
Managing aluminum phosphide poisonings
The Quadratic Equation http://cse.unl.edu/~ylu/raik283/notes/Resoureces/Quadratic_equation.htm