In this post, we will specifically look at some acids whose n-factor calculation can seem a little tricky. We will study the structure of these acids and then find their n-factor accordingly.
Example 3 – Consider boric acid (H_{3}BO_{3}).
Looking at the formula, one may conclude that the n-factor for this acid is 3 as there are three hydrogens present. However, the n-factor for this acid is 1. This is because out of the three hydrogens, there is only one replaceable hydrogen! Boric acid forms a complex ion with water and forms only one H^{+} ion as follows-
How can boron form such a complex? Why does it have a vacant p-orbital? To understand this, we first write its electronic configuration –
Boron , B (5) 1s^{2} 2s^{2} 2p^{1}
As seen above, boron has 3 valence electrons (shown in the red color box)- two in 2s and 2p orbitals. In an excited state, one electron from 2s orbital gets promoted to a 2p orbital and thus there are 3 unpaired electrons. However, there is one vacant 2p orbital that can accept a pair of electrons. Thus, boron can bear 4 OH^{–} groups on it. Boron has a negative charge due to these extra electrons.
∴ n-factor(n_{f}) of boric acid is 1.
M.wt of H_{3}BO_{3} = 3(1)+ 10.8 +3(16) = 3+11+48 = 62.
From the previous post, we know, equivalent weight = M.wt/ n_{f} .
∴equivalent weight = 62/1=62.
Example 4 – We shall now study three acids namely – H_{3}PO_{4} , H_{3}PO_{3} and H_{3}PO_{2. }The n-factor for these acids are as follows –
Acid |
n-factor |
H_{3}PO_{4} | 3 ,2 or 1 |
H_{3}PO_{3} |
2 or 1 |
H_{3}PO_{2} |
1 |
To understand why the n-factor varies, we need to draw the structures of these acids. Phosphorus is the central element in all these acids and it is pentavalent. The three structures are shown below –
Replaceable hydrogen → can form H^{+} ions on dissociation.
Non-replaceable hydrogen→ cannot form H^{+} ions on dissociation.
As seen in the figure above, the hydrogen atoms that are attached to oxygen(H) are replaceable. This is because oxygen is more electronegative than hydrogen. Therefore, oxygen pulls the electrons (that are shared between O and H) more towards itself. This allows the hydrogen to become replaceable. However, the hydrogens that are directly attached to phosphorous(H), are tightly bound. This is because the electronegativities of P and H are nearly similar.
Thus, the n-factors for three acids vary.
Calculations for H_{3}PO_{4 }–
n-factor of H_{3}PO_{4} is 3.
M.wt of H_{3}PO_{4} = 3(1)+ 31 +4(16) = 98g/mol
We know, equivalent weight = M.wt/ n_{f} .
∴Eq.Wt = 98/3=32.7g/eq.
Calculations for H_{3}PO_{3 }–
n-factor of H_{3}PO_{3} is 2.
M.wt of H_{3}PO_{3} = 3(1)+ 31 +3(16) = 82g/mol
We know, equivalent weight = M.Wt/ n_{f} .
∴Eq.Wt = 82/2=41g/eq.
Calculations for H_{3}PO_{2 }–
n-factor of H_{3}PO_{2} is 1.
Molecular mass of H_{3}PO_{3} = 3(1)+ 31 +2(16) = 66 g/mol
We know, equivalent weight = M.Wt/ n_{f} .
∴Eq.Wt = 66/1=66 g/eq.
We will continue to learn more about finding n-factor and calculating equivalent weights in the next post. Till then,
Be a perpetual student of life and keep learning…
Good Day!
Image source –
1)https://www.boricacid.net.au/product/boric-acid-powder-1kg/