Balance the chemical equation given below, and determine the number of milliliters of 0.00300 m phosphoric acid required to neutralize 25.00 ml of 0.00150 m calcium hydroxide. ca(oh)2(aq) + h3po4(aq) → ca3(po4)2(s) + h2o(l)

3 Ca(OH)2(aq) + 2 H3PO4(aq) → Ca3(PO4)2(s) + 6 H2O(l)

8.33 ml of phosphoric acid

Explanation:

To balance the equation, first add a 3 before Ca(OH)2, to balance Ca. Then, add a 2 before H3PO4, to balance P. Now you have 12 H on the left, so you have to add a 6 before H2O, to balance H. Finally, check Oxygen to notice that it is already balanced (there are 14 atoms at both sides).

3 Ca(OH)2(aq) + 2 H3PO4(aq) → Ca3(PO4)2(s) + 6 H2O(l)

In 0.00150 M Ca(OH)2 there are 0.0015 mol in 1 liter (or 1000 ml). Then in 25 ml, there are:

0.0015 mol / x mol = 1000 ml / 25 ml

x = 0.0015*25/1000

x = 3.75*10^(-5) mol of Ca(OH)2

From the balanced equation we know that 3 mol of Ca(OH)2 reacts with 2 mol of H3PO4, then 3.75*10^(-5) mol of Ca(OH)2 reacts with:

3 mol of Ca(OH)2 / 3.75*10^(-5) mol of Ca(OH)2 = 2 mol of H3PO4 / x mol of H3PO4

x = 2*3.75*10^(-5)/3

x = 2.5*10^(-5) mol of H3PO4

In 0.0030 M H3PO4 there are 0.003 mol in 1 liter (or 1000 ml). Then, 2.5*10^(-5) mol of H3PO4:

0.003 mol / 2.5*10^(-5) mol = 1000 ml / x ml

x = 1000*2.5*10^(-5)/0.003

x = 8.33 ml of H3PO4

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.