# How do we calculate the basicity of H3PO4

## pKs and pKb values

If we consider two different acids with the same concentration, they can still have different pH values. Why is that?

This is because some acids are stronger than others, so they can give up their protons more easily. Of course, there is also a variable here that tells us about it.

It is the acid constant KS or the pKS value. The acid constant, just like the ion product of water, is derived from the MWG and is a measure of how “gladly” an acid gives off a proton to water.

If we apply the MWG again here, we get the following equilibrium constant:

\ begin {align *}

K = \ frac {c ({H_3O ^ {+}}) \ cdot c ({A ^ {-}})} {c ({HA}) \ cdot c ({H_2O})}

\ end {align *}

Here too, as with the ionic product of water, we can consider the concentration of the water to be approximately constant (since there is so much available that the concentration of the water hardly changes with the small amount of acid) and offset it with the equilibrium constant:

\ begin {align *}

\ begin {array} {crcll}

& K & = & \ frac {c ({H_3O ^ {+}}) \ cdot c ({A ^ {-}})} {c ({HA}) \ cdot c ({H_2O})} & | \ cdot c ({H_2O}) \ \

\ Leftrightarrow & K \ cdot c ({H_2O}) & = & \ frac {c ({H_3O ^ {+}}) \ cdot c ({A ^ {-}})} {c ({HA})} & | \ \ text {Set} K \ cdot c ({H_2O}) = K_ \ text {S} \ \

\ Rightarrow & K_ \ text {S} & = & \ frac {c ({H_3O ^ {+}}) \ cdot c ({A ^ {-}})} {c ({HA})} & \ text { Acid constant}

\ end {array}

\ end {align *}

If we now apply the negative decadic logarithm to the acid constant, as in the pH value calculation, we get the pKa value. In the same way we can give the pKB value as a measure of the strength of a base. The value is a measure of how “gladly” a base takes up a proton from water.

With the help of the pKS and pKB values, we can now compare different acids or bases with regard to their strength.

- The smaller the pKa value, the stronger the acid.
- The smaller the pKB value, the stronger the base.

With the help of the following table we can compare some important acids and bases.

\ begin {array} {| c | c | c | c |} \ hline

{pK_ \ textbf {S}} &

{\ textbf {acid}} &

{\ textbf {Corresponding Base}} &

{pK_ \ textbf {B}} \

\ hline

{\ text {Complete proton donation}} &

{HCIO_4} &

{CIO_4 ^ {-}} &

{\ text {No proton uptake}} \

\ hline

{\ text {Complete proton donation}} &

{HI} &

{I ^ {-}} &

{\ text {No proton uptake}} \

\ hline

{\ text {Complete proton donation}} &

{HCI} &

{CI ^ {-}} &

{\ text {No proton uptake}} \

\ hline

{\ text {Complete proton donation}} &

{H_2SO_4} &

{HSO_4 ^ {-}} &

{\ text {No proton uptake}} \

\ hline

-1,74 &

{H_3O ^ {+}} &

{H_2O} &

15,74 \\

\ hline

-1,32 &

{HNO_3} &

{NO_3 ^ {-}} &

15,32 \\

\ hline

1,92 &

{HSO_4 ^ {-}} &

{SO_4 ^ {2-}} &

12,08 \\

\ hline

2,13 &

{H_3PO_4} &

{H_2PO_4 ^ {-}} &

11,87 \\

\ hline

3,14 &

{HF} &

{F ^ {-}} &

10,86 \\

\ hline

3,35 &

{HNO_2} &

{NO_2 ^ {-}} &

10,65 \\

\ hline

3,75 &

{HCOOH} &

{HCOO ^ {-}} &

10,25 \\

\ hline

4,75 &

{CH_3COOH} &

{CH_3COO ^ {-}} &

9,25 \\

\ hline

6,52 &

{H_2CO_3} &

{HCO_3 ^ {-}} &

7,48 \\

\ hline

6,92 &

{H_2S} &

{HS ^ {-}} &

7,08 \\

\ hline

7,00 &

{HSO_3 ^ {-}} &

{SO_3 ^ {2-}} &

7,00 \\

\ hline

7,20 &

{H_2PO_4 ^ {-}} &

{HPO_4 ^ {2-}} &

6,80 \\

\ hline

9,25 &

{NH_4 ^ {+}} &

{NH_3} &

4,75 \\

\ hline

9,40 &

{HCN} &

{CN ^ {-}} &

4,60 \\

\ hline

10,40 &

{HCO_3 ^ {-}} &

{CO_3 ^ {2-}} &

3,60 \\

\ hline

12,36 &

{HPO_4 ^ {2-}} &

{PO_4 ^ {2-}} &

1,64 \\

\ hline

13,00 &

{HS ^ {-}} &

{S ^ {2-}} &

1,00 \\

\ hline

15,74 &

{H_2O} &

{OH ^ {-}} &

-1,74 \\

\ hline

{\ text {No proton release}} &

{C_2H_5OH} &

{CC_2H_5O ^ {-}} &

{\ text {Complete proton uptake}} \

\ hline

{\ text {No proton release}} &

{NH_3} &

{NH_2 ^ {-}} &

{\ text {Complete proton uptake}} \

\ hline

{\ text {No proton release}} &

{OH ^ {-}} &

{O ^ {2-}} &

{\ text {Complete proton uptake}} \

\ hline

{\ text {No proton release}} &

{H_2} &

{H ^ {-}} &

{\ text {Complete proton uptake}} \

\ hline

\ end {array}

The top four acids give up their protons completely, that is, a complete protolysis takes place (we can also say that the acid dissociates completely). The equilibrium is therefore on the right. The further you go down in the table, the higher the pKa values and therefore the weaker the acids. The table also shows the corresponding acid-base pairs.

It becomes clear that the corresponding base is weaker the stronger an acid is. The reverse is also true: the stronger a base, the weaker the corresponding acid.

There is even a very specific relationship between the pKa value of an acid and the pKB value of the corresponding base:

If we apply the negative decadic logarithm here, we get:

\ begin {align *}

pK_ \ text {S} + pK_ \ text {B} = 14

\ end {align *}

If the pKa value of an acid is known, we can use the formula to calculate the pKB value of the corresponding base.

If we look at the bases in the table, we can see that the bottom four bases are very strong, as they take up protons completely. This means that all base particles take up a proton.

Now we can ask ourselves why it is so important whether an acid or base is strong or weak. This is especially important when we want to calculate the pH of a solution. To do this, we have to be able to classify the pKS and pKB values correctly.

In order to correctly classify the strength of acids and bases, the following values are relevant for us:

\ begin {align *}

\ begin {array} {ll | ll}

pK_ \ text {S} <1.5 & strong \ acid & \ quad pK_ \ text {B} <1.5 & strong \ base \

1.5

\ end {array}

\ end {align *}

### pH calculation of acids and bases of different strengths

### Strong acids

If the pKa value of an acid is less than 1.5, it is a strong acid. Then we can assume that the protolysis is almost complete. Therefore, all acid particles give their protons to the water, whereby the concentration of oxonium ions is approximately the same as the initial acid concentration.

So we can calculate the pH value as follows:

\ begin {align *}

pH = - \ lg \ left (c ({H_3O ^ {+}}) \ right) = - \ lg \ left (c_0 ({HA}) \ right)

\ end {align *}

### Weak acids

In the case of weak acids, only very few acid particles give off their protons to the water. Therefore, the concentration of the acid hardly changes as a result of the protolysis, i.e. c (HA) ≈ c0 (HA). The concentration of the oxonium ions is just as large as the concentration of the corresponding base, since for each acid particle that gives off a proton, an oxonium ion and a particle of the corresponding base are produced, i.e. c (H3O +) = c (A−). If you put these two relationships into the MWG, you get the following equation:

\ begin {align *}

\ begin {array} {crcll}

& K_ \ text {S} & = & \ frac {c ({A ^ {-}}) \ cdot c ({H_3O ^ {+}})} {c_0 ({HA})} & \

\ Leftrightarrow & K_ \ text {S} & = & \ frac {c ({H_3O ^ {+}}) ^ 2} {c_0 ({HA})} & | \ cdot c_0 ({HA})

\ end {array}

\ end {align *}

To find the pH we need the concentration of the oxonium ions, so we solve the equation for the concentration:

\ begin {align *}

\ begin {array} {crcll}

& c ({H_3O ^ {+}}) ^ 2 & = & K_ \ text {S} \ cdot c_0 ({HA}) & | \ \ sqrt {} \

\ Leftrightarrow & c ({H_3O ^ {+}}) & = & \ sqrt {K_ \ text {S} \ cdot c_0 ({HA})} & | \ \ text {neg. Apply logarithm} \

\ Rightarrow & \ text {pH} & = & - \ lg \ left (\ sqrt {K_ \ text {S} \ cdot c_0 ({HA})} \ right) & \

& & = & - \ frac {1} {2} \ cdot \ lg \ left (K_ \ text {S} \ cdot c_0 ({HA}) \ right) &

\ end {array}

\ end {align *}

So we can calculate the pH value as follows:

\ begin {align *}

pH = \ frac {1} {2} \ cdot \ left (\ text {p} K_ \ text {S} - lg \ (c_0 ({HA}) \ right))

\ end {align *}

### Strong bases

In the case of strong bases, we can basically use the same formula as in the case of strong acids. Attention: If we calculate the negative decadic logarithm of the initial concentration of the base, we do not get the pH value, but the pOH value, since the reaction of a base with water does not produce oxonium ions, but hydroxide ions. But this is not a problem, because with the help of the pOH value we can calculate the pH value, namely again with the help of the ion product of the water. Because this results in the following relationship between pH and pOH:

\ begin {align *}

\ text {pH} + \ text {pOH} = 14

\ end {align *}

The pH value of strong bases can be calculated as follows:

\ begin {align *}

pH = 14 + \ lg \ left (c_0 ({A ^ {-}}) \ right)

\ end {align *}

### Weak bases

Even with weak bases, the pH value calculation works with the same formula as with weak acids, only that we have to be careful that the pOH value is calculated first. From this we calculate the pH value. This results in the following formula:

\ begin {align *}

pH = 14 - \ frac {1} {2} \ cdot \ left (\ text {p} K_ \ text {S} - c_0 ({HA}) \ right)

\ end {align *}

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