Question #6d6d4

2 Answers
Oct 31, 2015

Answer:

#P(V)#; the maximum oxidation number.

Explanation:

We have #H_2PO_4^-#. The sum of the oxidation numbers must give the charge on the ion, which is #-1#. Oxygen generally has an oxidation state of #-II# in its compounds, and it does here; hydrogen generally has an oxidation state of #+1#. Is the answer correct? Don't assume I have done my arithmetic correctly; I have trouble tying my shoelaces!

Would the answer change if I had the parent acid, phosphoric acid, #H_3PO_4#? Why or why not?

Oct 31, 2015

Answer:

+5

Explanation:

Okay, with determining the oxidation states there are some rules. This would help you to not get confuse.

(1) The total charge of a stable compound is always equal to zero (meaning no charge).

For example, the #H_2O# molecule exists as a neutrally charged substance.

(2) If the substance is an ion (either there is a positive or negative charge) the total oxidation state of the ion is the charge (i.e. oxidation state of #NO_3^"-1"# ion is -1).

(3) All elements from Group 1A has an oxidation state of +1 (e.g. #Na^"+1"#, #Li^"+1"#). All Group 2A and 3A elements have an oxidation state of +2 and +3, respectively. (e.g. #Ca^"2+"#, #Mg^"2+"#, #Al^"3+"#)

(4) Oxygen always have a charge -2 except for peroxide ion (#O_2^"2-"#) which has a charge of -1.

(5) Hydrogen always have a charge of +1 if it is bonded with a non-metal (as in the case of #HCl#) and always have a -1 charge if it is bonded with a metal (as in #AlH_3#).

So let's try solving your problem, the oxidation state of #P# in the substance #KH_2PO_4#.

Based on rule 1, the whole substance has an oxidation state of zero.

#KH_2PO_4# = 0

Based on rules 3 and 5, the oxidation states of #K# and #H# atoms are both +1 (since there no metal atom in this substance).

(+1) + [(+1)(2)] + #PO_4# = 0

Notice that since #H# atoms have a subscript, I multiply its oxidation state by 2.

Next, based on rule 4 the #O# atom has a charge of -2.

(+1) + [(+1)(2)] + #color (red) x# + [(-2) (4)] = 0 where #color (red) x# is the oxidation state of #P# atom

Notice again that since #O# atom has a subscript, I multiplied the oxidation state by 4. Now, we are ready to solve for #x#.

(+1) + [(+1)(2)] + #color (red) x# + [(-2) (4)] = 0

(+1) + (+2) + #color (red) x# + (-8) = 0

(+3) + #color (red) x# + (-8) = 0

#color (red) x# + (-5) = 0

#color (red) x# = +5

Therefore, the oxidation state of #P# atom is +5 or #P^"5+"#.