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Question: The degree of ionization of a \(0.1M\) bromoacetic acid solution is 0.132. Calculate the \(pH\) of t...

The degree of ionization of a 0.1M0.1M bromoacetic acid solution is 0.132. Calculate the pHpH of the solution and the pKap{K_a} , of a bromoacetic acid.
pHpH of solution is = 1.881.88
pKap{K_a} of bromo acetic acid = 2.692.69

Explanation

Solution

pHpH is a measure of hydrogen ion concentration, [H+][{H^ + }] ,in an aqueous (water) solution. The pHpH scale ranges from 0140 - 14 . A low pHpH value indicates acidity, a pHpH of 77 is neutral, and a high pHpH value indicates alkalinity.
The formulas to calculate pHpH is:
pH=log[H+]pH = - \log [{H^ + }]
At 250C{25^0}C :
pH+pOH=14pH + pOH = 14

Ka{K_a}pKap{K_a}
It is the acid dissociation constant.It is the log- \log of this constant.
Describes the degree of ionization.Also describes the degree of ionisation.
True indicator of acid strength.Reason: Addition of water does not change the equilibrium.True indicator of acid strength.Reason: Addition of water does not change the equilibrium.

The acid dissociation constants are usually expressed in terms of l1mol{l^{ - 1}}mol .
Acids and bases dissociate according to general equations:
HA+H2OA+H3O+HA + {H_2}O \to {A^ - } + {H_3}{O^ + }
Ka=[H+][A][HA]{K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}
pKa=logKap{K_a} = - \log {K_a}

At half the equivalence point,
pH=pKapH = p{K_a} pH=pKa=logKpH = p{K_a} = - \log K

Complete step by step answer:
CH2BrCOOHCH2BrCOO+H+C{H_2}BrCOOH \to C{H_2}BrCO{O^ - } + {H^ + }

Initial concentrationCC0000
Concentration at equilibriumCCαC - C\alpha CαC\alpha CαC\alpha

So, the acid dissociation constant will be:
Ka=Cα×CαC(1α)Cα21αCα2{K_a} = \dfrac{{C\alpha \times C\alpha }}{{C(1 - \alpha )}} \Rightarrow \dfrac{{C{\alpha ^2}}}{{1 - \alpha }} \simeq C{\alpha ^2}
 0.1×0.132 1.74×103 pKa=log(1.74×103) 2.76  \ \Rightarrow 0.1 \times 0.132 \\\ \Rightarrow 1.74 \times {10^{ - 3}} \\\ p{K_a} = - \log (1.74 \times {10^{ - 3}}) \\\ \Rightarrow 2.76 \\\ \
From the above values, we can find the concentration of hydrogen ion as:
 [H+]=Cα0.1×0.132 1.32×102M  \ [{H^ + }] = C\alpha \Rightarrow 0.1 \times 0.132 \\\ \Rightarrow 1.32 \times {10^{ - 2}}M \\\ \
So, the final calculation of pHpH will be:
 pH=log(1.32×102) 1.88  \ pH = - \log (1.32 \times {10^{ - 2}}) \\\ \Rightarrow 1.88 \\\ \

Note:
A large Ka{K_a} value indicates a strong acid because it means the acid is largely dissociated into its ions. A large Ka{K_a} value also means the formation of products in the reaction is favored. A small Ka{K_a} value means little of the acid dissociates, so you have a weak acid. The Ka{K_a} value for most weak acids ranges from 1021014{10^{ - 2}} - {10^{ - 14}} .
The pKap{K_a} gives the same information, just in a different way. The smaller the value of pKap{K_a} , the stronger the acid. Weak acids have a pKap{K_a} ranging from 2142 - 14
Another important point is pIpI .This is the isoelectric point. It is at which protein molecule is electrically neutral that is having no net electric.